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Lessons in Horology 


BY 

Jules Grossmann 

i 1 

Director of the Horological School, of Locle, Switzerland 

AND 


Hermann Grossmann 


Director of the Horological and Electro-Mechanical 


School, of Neuehatel, Switzerland 


AUTHORIZED TRANSLATION 
By JAMES ALLAN, JR. 

V 

of Charleston, S. C. 


Former Pupil of the Horological School, of Locle, Switzerland 


VOLUME I 

The Principles of Cosmography and Mechanics Relat 
ing to the Measurement of Time—Motive Force, 
Mainsprings, Trains, Gearings, etc. 

WITH OVER 100 ILLUSTRATIONS 




PUBLISHED BY 

THE KEYSTONE 

THE ORGAN OF THE JEWELRY AND OPTICAL TRADES 
19 TH & Brown Sts., Philadelphia, U.S.A. 

1905 

All flights f \served 







,G-*7 


LIBRARY of CONGRESS 
Two Copies Received 

DEC 23 1905 

Copyright Entry 

(9 c /■ / 3 , /90 6~ 

CUSS CL XXc. No. 

/ XX 6 / L 

COPY B. 


Copyright, 1905 , by B. Thorpe 
Publisher of The Keystone 





PREFACE BY THE AUTHORS 


O one ignores the fact that horology has attained a highly 



x ^ prominent position among the mechanical arts during the 
past quarter of a century, and that this fact is due to the rapid 
progress of science, industry and commerce in our epoch. If it 
sufficed formerly to know the time, approximately, and to make 
use of the indications of the sun dials, town clocks with single hands 
and other primitive instruments, modern times, with their multi¬ 
plied requirements, have rapidly despoiled us of this ancient 
simplicity; they demand of us an exact marking of every instant, 
which only modern horary instruments permit us to attain. 

The magnificent chronometers, whose superior timekeeping 
we admire to-day, are the product of the two-fold effort; first, of 
the theorist who, by his calculations, determines all the principles; 
then of the practical workman who faithfully follows these in the 
execution of his work. 

In the period in which we live we cannot believe that we have 
yet attained the highest point of precision, but the results are 
already so brilliant that the mind now asks the question whether, 
before going farther into the technical domain, it would not be 
better first to bring to perfection the means of observation and of 
rating, which we now invariably employ with a degree of uncertainty. 
The magnificent instruments to which we have alluded are, more¬ 
over, still exceptions; they are very expensive as yet. So the 
most practical object of the technical study of horology is to 
approach, as nearly as possible in public timepieces, the results of 
the precision chronometer, at least so far as concerns the exactness 
of timekeeping. This purpose will surely be attained when horology, 
seconded by the admirable resources of mechanics, will entirely 
cease to be an art, too often empirical, and become a purely 
mechanical science. 

It has long been believed that the theory of horology formed a 
science by itself, independent of general mechanics, and for a long 
time the watchmaker would not listen to anything about mechanics, 



VI 


Preface 


pretending that it was impossible to apply its data to the minute 
pieces which compose the mechanism of a pocket watch. This 
assertion was often apparently sustained by practical results, and 
frequently the purely mechanical data appeared as if they could not 
be applied to horology. But this conclusion, let us hasten to say, 
was false; for the reason that the mathematical formulas employed 
in mechanics often require less development than when they are 
applied to horology. In the first case many of the terms could be 
neglected which in the second would become important. We must 
not be astonished, either, if the results are not always what we seek. 
Let us take an example: Would one really dare to pretend that 
the laws of friction established by Coulomb, are inexact because it 
is very difficult in horology to separate friction proper from the 
influence of adhesion produced by the oil or other lubricating 
material ? This second factor, which may often be omitted in large 
mechanics, becomes, we know, an important factor in horology. 
The work which we present to the English-speaking watchmakers 
is written by watchmakers and for watchmakers, and with the idea 
that horology and mechanics are twin sisters, and that the same 
laws and the same rules control both. 

This work is the fruit of long experience in the domain of 
professional instruction in horological schools. We have endeavored 
to avoid speculations purely theoretical, as well as long descriptive 
explanations, which belong to books suited to the general public. 
If the solution of some problems cannot, in our estimation, be 
accomplished without the aid of higher mathematics, because of 
the precision required to attain the desired end, it must be noted 
that these questions can generally be put aside by those persons to 
whom the subtleties of mathematical analysis are unfamiliar. 

It is sufficient then to recognize the fact that the calculations 
have been made to verify the deductions and to make use of the 
conclusions which may be drawn from them. 

We are also obliged to grade the difficulties of calculation so 
that they are presented in proportion to the development of the 
mathematical knowledge of the persons who undertake the study, 
and we follow each problem with at least one numerical application. 
When it is possible we give also together with a complicated 
solution, another similar to it, but more simple. 

Our plan comprises, first, a short introduction on the principles 
of cosmography and mechanics having relation to the measuring of 


Preface 


vu 


time. Then follow chapters which are devoted to the study of 
motive forces produced by the weight and the barrel spring, the 
calculations of trains and the theory of gearings. Then chapters 
on escapements, and finally the theory of adjusting and regulating 
forms an important part of the work, and will be treated with all 
the exactness due the subject. We will close this exposition of 
the theory by a treatise on the compensation of chronometers. 

We hope that this work will contribute its share towards 
forming a generation of capable and educated horologists who can 
assist in the development of the fascinating industry of horology. 

We owe a just tribute of appreciation to The Keystone, which 
has undertaken the publication of this work in the English language, 
and to James Allan, Jr., of Charleston, S. C., former pupil of the 
Locle Horological School, who has so well performed the work of 
translation. 

Jules Grossmann, 

Locle, Switzerland. 

Hermann Grossmann, 

NeuchateJ Switzerland. 


4 






TABLE OF CONTENTS. 


Paragraphs. Page*. 

Preface.3 

INTRODUCTION. 

I. General Principles of Cosmography Relating 

to Horology. 

1 Principles of the measurement of time. 9 

3 Units of time. Sidereal day. Solar day. 9 

7 True time. Mean time.n 

9 Laying out of a meridian line.12 

13 Determination of the position of a point on the terrestrial sphere . 14 

II. General Principles of Mechanics. 

19 Forces.19 

27 Law of inertia.20 

28 Definition of mechanics.20 

29 Motion . 21 

34 Rotary motion.22 

35 Mass of a body.22 

Work of a force.23 

37 Definition.23 

38 Work of a force tangent to a wheel.24 

39 Unit of work.24 

40 Active power.24 

Moment of a force.25 

43 Lever.26 

Transmission of work in machines. 27 


CHAPTER I. 

General Functions of Clocks and Watches. 

The oscillations of the pendulum and their relation to the motive 

force.3 2 


(l). 
























2 


Lessons in Horology. 


Paragraphs. Pages. 

The oscillations of the balance and their relation to the motive 

force .35 

Wheel-work.—Its purpose in the mechanism of clocks and watches . 37 

Escapements.39 


CHAPTER II. 

Maintaining cr Motive Forces. 

The weight as a motive force. 40 

The barrel spring as a motive force. 43 

81 Measurement of the force of a spring. 43 

Theoretical study of the moment of a spring’s force . . . . . . . 45 

86 Coefficient of elasticity. 45 

91 Variation of the coefficient of elasticity. 47 

92 Values of the coefficient of elasticity E . 48 

93 Limit of elasticity. 49 

94 Moment of the elastic force of a spring subjected to a flexion . . 49 

100 Inequality of the elastic force of the spring. 56 

104 Length of the spring. 58 

Development of a spring. 59 

hi Diameter of the hub. 64 

Work produced by spring. 66 

The fusee. 67 

115 Calculation of the variable radius of the fusee’s helix. 69 

119 Uniformity of the force of the spring in fusee watches. 74 

Stop-work. 78 

128 Geometrical construction of the Maltese cross stop-work .... 79 


CHAPTER III. 

Wheel-Work. 

129 Purposes of wheel-work. 82 

Calculations of trains... 82 

130 Calculations of the number of turns. 82 

133 Calculation of the number of oscillations of the balance. 84 

136 Calculations of the numbers of teeth. 86 

Problems relative to the preceding questions. 89 





























Table of Contents. 


3 


Paragraphs. Pages. 

141-145 Numbers of turns.89-90 

146 Numbers of oscillations of the balance. 90 

147-149 Numbers of teeth.91-92 

150 Numbers of teeth of the minute wheels. Description of this 

mechanism. 93 

I53~ I 54 Numbers of teeth of an astronomical clock. 97-99 

I 55~ I 5S Numbers of teeth of lost mobiles.100-102 

159 Indicator of the development of the spring in fusee timepieces . . 102 

160 Simple calendar watches.104 

161 Decimal watches.106 

164 Calculation of numbers comprising the teeth-ranges of the wheels 

of a watch with independent second hand.109 

168-169 Wheel-work of the stem-winding mechanism.112-113 

169 a Calculation of the train in a watch of the Roskopf type .... 114 


CHAPTER IV. 

Gearings. 

170 Definition.117 

173 Practical examination of a gearing.118 

First—Distance of the centers.119 

176 Primitive radii.119 

179 Applications.121 

185 Calculation of the primitive radii.125 

193 Application of the theory of primitive radii to the escapements . . 129 

Second—Shape of the teeth and leaves . ..130 

195 General study of the transmission of force in gearings.130 

Determination of the forms of contact in gearings.136 

209 First—Graphic method. Exterior gearing.137 

210 Interior gearing. 

211 Second—Method of the envelopes.14° 

219 Evolvents of circle gearings.146 

220 Third—Roller method. 

221 Flank gearings. x 5 ° 

226 a Determination of the profile of a tooth corresponding to a profile 

chosen arbitrarily. *54 

22 6 b Gearings by the evolvent of a circle. 155 

Teeth-range. *56 

Third—Total diameters.*. *59 
































4 


Lessons in Horology. 


Paragraphs. Pages. 

Cycloid. *59 

235 Definition. x 59 

236 Drawing of the cycloid. *59 

237 Drawing of the cycloid of a continuous movement.160 

238 Normal and tangent to the cycloid.161 

239 Evolute and radius of curvature of the cycloid.161 

240 Length of the cycloid.162 

Epicycloid.163 

241 Definition.163 

242 Drawing of the epicycloid .163 

243 Drawing of the epicycloid of a continuous movement.164 

244 To draw a normal, then a tanget to the epicycloid.164 

245 Evolute and radius of curvature of the epicycloid.165 

250 Length of the epicycloid.169 

2 5 I_2 53 Applications.170-171 

Relation of the radius vector to the angle formed by the variable 

radius vector and the initial radius vector.171 

Table showing the angle traversed by the pinion of several ordi¬ 
nary gearings during the contact of a tooth of the wheel with 
the leaf of this pinion.175 

Calculation of the total radius of the wheel.176 

Form of the excess of the pinion leaf in a flank gearing.176 

268 Radius of curvature of an elipse.181 

270 Total radius of the pinion.184 

Graphical construction ©f gearings.185 

Practical applications of the theory of gearings.188 

276 The proportional compass and its use.190 

Table for using the proportional compass .195 

284 Verification of a proportional compass.196 

285 Determination of the distance of the centers of a gearing by 

means of the proportional compass and of a depthing tool . . 196 

286 The proportional compass and stem-winding wheel gearings . . . 197 

286 Gearing of the crown wheel in the ratchet wheel.198 

287 Gearing of the winding pinion in the crown teeth of the contrate 

wheel.199 

288 Gearing of the sliding pinion and of the small setting wheel . . . 200 

289 Gearings of the dial wheels.201 

Various calculations relative to gearings.202 

Conical gearings.214 

315 Form of the teeth.216 

































Table of Contents. 


5 


Paragraphs. Pages. 

318 Construction of conical gearings. 218 

Defects which present themselves in these gearings.222 

Passive resistances in gearings.227 

326 General ideas .227 

Friction.228 

330 The two kinds of friction.229 

332 Laws of friction.229 

333 Experimental determination of the force of friction.230 

335 Table of the coefficients of friction.231 

336 Work of friction.232 

337 Angle of friction.233 

Calculation of the friction in gearings.234 

339 Friction of the teeth ranges.234 

346 Friction before and after the line of centers.238 

348 1 st. The wheel drives the pinion after the line of centers ... 239 

349 2d. The pinion drives the wheel before the line of centers . . . 241 

350 3d. The wheel drives the pinion before the line of centers . . . 242 

351 4th. The pinion drives the wheel after the line of centers .... 244 

352 Recapitulation of the preceding calculations.244 

Calculations of the friction of pivots.245 

354 Work absorbed by the friction of the plane surface of the shoulder 

of a pivot .245 

356 Work absorbed by the friction of the cylindrical surface of a pivot 247 

357 Determination of the lateral pressure received by the pivots of the 

mobiles in a train.247 

Influence of the oil.253 

Application of the theory of gearings.255 

Functions of the heart in chronographs.255 























LESSONS IN HOROLOGY 













LESSONS IN HOROLOGY. 


COURSE IN MECHANICS AS APPLIED TO CHRONOMETRY. 


I. Introduction—General Principles of Cosmography Relating to Horology. 

1. Principles of the measurement of time. Cosmography is a 
science which has for its object the study of the different celestial 
phenomena as they are given to us by observation and calculation ; 
it comprehends also the study of the principles which are the basis 
of the measuring of time. 

When one proposes to measure a length, surface, volume or 
weight of any kind one chooses arbitrarily a unit of length, surface, 
volume or weight with which one compares the object to be 
measured, noting exactly the number of times that this unit is 
contained therein. 

When the purpose is to measure the intervals of time, it is no 
longer possible to make use of an analogical method. But to 
effect this operation one is obliged to determine the space traversed 
by a body animated with a uniform or periodically uniform motion 
(31). In the first place one concludes that the intervals of time 
are proportionate to the spaces traversed by the body considered. 

2 . It is necessary then to admit that all measurements of time 
must be deduced from the observation of a regular movement. 
Thus, formerly, one determined the fraction of time, more or less 
great, by the running of sand in the “hour glass,” or of water in 
the “ clepsydra.” Now time is measured in clocks and in watches 
by the periodically uniform movement of the pendulum or balance 
wheel. 

3. Units of time. Sidereal day. Solar day. For the determi¬ 
nation of the unit of time it is necessary to choose the most uniform 
movement possible, a movement whose speed must be the same 

' to-day, to-morrow, in a year or in an indefinitely prolonged period. 
Such a movement filling this condition absolutely, is the rotation of 
the earth on its axis ; no cause or effect whatever could increase or 
diminish it. We have positive proofs that this movement is the 
same to-day as it was in the time of Hipparchus, an ancient 
astronomer of the school of Alexandria who lived two centuries 


9 



io Lessons in Horology. 

before Jesus Christ. We can assure ourselves by the calculation 
of the eclipses, that the length of one of these rotations is the same 
to-day as in the time of that astronomer within of a second. 
This movement has then been chosen because of its great regu¬ 
larity as the basis for the measuring of time. 

The duration of a complete rotation is the unit, and is called 
a day. 

4. In order to determine with exactitude the commencement 
and the end of this movement, it is necessary to choose a point of 
repose outside of the earth, and for this purpose a fixed star or the 
sun is taken. Let us remark that the result differs according as we 
take one or the other of these two points. The following demons¬ 
tration will explain the reason. 

5. We know that the earth not only turns on its axis, but that 
it has also a simultaneous movement around the sun. Let us take, 
then T and T' , Fig. i, as the two positions which the earth 
occupies in its orbit at the commencement and at the end of one of 
its diurnal rotations. In the first of these positions, a is a point on 
its surface from which can be seen at this instant the center of the 
sun S, in an imaginary plane passing through the two poles and the 
point considered a , this plane is the meridian plane. At the end 
of a certain time, the earth has traveled in its orbit to the position 
T f and the point a arrives at a’ in such a manner that the line 
T' a ' is parallel to T a. The earth will then have accomplished 
one rotation on its axis and all of its parts will have, with relation 
to the fixed stars, the same positions that they had at T. 

The time during which this rotation is accomplished is called 
a sidereal day. 

But from the point a' in the position T' one could not see the 
sun in the meridian plane ; in order that the observer placed at a' 
could perceive it anew in this plane, it would be necessary for the 
point a' to be removed to b in traversing the arc a! b. The solar 
day, that is to say, the time which elapses between two consecutive 
passages of the sun to the meridian plane is then longer than the 
sidereal day. In dividing the solar day into 24 hours, the hour 
into 60 minutes and the minutes into 60 seconds, the sidereal day 
counts only 23 hours, 56 minutes, 4.09 seconds ; the sidereal day is 
then shorter than the solar day 3 minutes, 55.91 seconds. 

If we divide, on the other hand, the sidereal day into 24 hours, 
the solar day will count 24 hours, 3 minutes, 56.55 seconds. This 


General Principles of Cosmography. n 

value of the solar day, variable from the sidereal day, is only a 
mean value. 

6 . The time that the earth takes to traverse its orbit, that is to 
say a year, contains exactly one sidereal day more than the solar days. 

7. True time. Mean time. The curve that the earth describes 



around the sun is an ellipse of which that star occupies one of the 
foci. Our planet does not traverse this curve with a uniform speed, 
it moves more rapidly when it is nearest the sun and more slowly 
when it is farthest away. The arcs traversed by the earth in one 
day are not then the same length during all the year. There 
results an irregularity in the duration of the solar day, the solar 
day is longest when the earth goes fastest, and it is shortest when 



12 


Lessons in Horology. 


its movement of translation is slowest. Another cause which still 
increases this irregularity is due to the fact that the axis of the earth 
is not perpendicular to the plane of the orbit that it traverses around 
the sun (plane of the ecliptic). 

The length of the solar day can vary in 24 hours as much as 
30 seconds, plus or minus. Thus the solar day with its diurnal 
variation of length does not fill at all the conditions desired for the 
measuring of time, the unit adopted must be of fixed value, so that 
our horological instruments, all based on a uniform movement, 
may follow their regular running without necessitating perpetual 
resetting. We fall then into a difficulty, since naturally the sun 
should measure the time for us, while in reality its unequal move¬ 
ment does not lend itself to this measuring. The difficulty has 
been adjusted as follows. We divide the total duration of the year 
by the number of solar days that it contains ; the quotient will 
be a mean value, shorter than the solar days of greatest length and 
longer than the solar days of least duration. It will be, moreover, 
almost equal to certain days between them. This mean value is 
called mean time. We call, on the contrary, true time the direct 
interval of time elapsing between two successive passages of the 
sun across the meridian. The difference, plus or minus, between 
true time and mean time can amount to as much as 17 minutes. 

8 . The equation Of time is the value that must be added to or 
subtracted from the true solar day to obtain the mean solar day. 
The year book of the Bureau of Longitudes announces each year in 
a calendar the result of the equation of time, and gives in a 
column entitled “Mean time at true noon” what a chronometer, 
regulated on mean time, should indicate at the exact moment 
of noon. The equation of time is nothing or almost that, four 
times a year—the 15th of April, 15th of June, 31st of August and 
25th of December, while it attains its greatest value between the 
10th and 12th of February and the first days of November. 

9. Laying out Of a meridian line. We already have an idea of 
the importance of the meridian plane in the determination of the 
length of a rotation of the earth on its axis. 

Let us see now how we can proceed to establish the direction 
of a meridian line, that is to say, of the trace of the meridian plane 
on the surface of the earth. Among the several methods known let us 
choose the following, which recommends itself on account of its ex¬ 
treme simplicity, and which does not require instruments of precision. 


General Principles of Cosmography. 


13 


On a horizontal plane, conveniently placed, we fix a vertical 
style ; from its foot, O, Fig. 2, we describe on this plane several 
concentric circles of any size, such as m n, m'n ’, etc. Let us mark 
on these circumferences the points A, B , C etc., where the extremity 
of the shadow of the pin reaches before midday. In the afternoon 
renew the operation by indicating in the same manner the points 
A\ B\ C\ etc. We connect the points marked on the same cir¬ 
cumference by a straight line, and we obtain thus as many straight 
lines as circles, and they are parallel to each other. The perpen¬ 
dicular laid off from the center O , on these straight lines, will be 


N 



the meridian line sought. Since the shadow cast by such a style 
is never very distinct, we will reach a greater accuracy by finishing 
the extremity of the pin with a metallic plate, in which we pierce a 
fine hole. We indicate then the center of the image of the sun on 
each of the circumferences, as previously done. In the above con¬ 
struction a single arc sufficed, but it is preferable to employ several 
which mutually control each other, the middle of the straight lines, 
AA\ BB\ CO , etc., should be found with the center O on the 
same perpendicular O N 

10. When the middle of the small image of the sun is found on 
the meridian it is nearest the foot of the style and consequently the 
sun is at its greatest height; it is then exactly midday. To obtain 
the mean hour we must consult a table of equations and add or 











14 


Lessons in Horology. 


subtract, according to the season of the year, the correction indi¬ 
cated for this day. Generally, as we have said, these tables indicate 
the “mean time at noon,” thus for the 17th of November, 1893, 
we find for example indicated : 

Mean time at noon 11 h. 45 m. 10 s. the clock or watch should 
then mark 11 hours, 45 minutes and 10 seconds when the middle of 
the small image of the sun is projected on the meridian line. 

11. The Meridian Glass employed in the observatories is nothing 
else than a meridian line determined with the greatest exactness. 
It is generally a glass of sufficiently large size which can only move 
in the meridian plane, and is divided by it into two symmetrical 
parts. It is supported on two immovable pillars by means of trun¬ 
nions, which permit it to take all the positions possible around its 
axis of rotation. It can then be used to observe the passage on the 
meridian of all the visible stars above the horizon. The sensitive¬ 
ness of the instrument is moreover augmented by the magnifying 
power of the glasses employed. 

Since such an instrument cannot be transported, we have 
recourse to other instruments in order to determine the hour at any 
locality, on the sea for example. The one generally used is the 
sextant , but its employment is complicated. 

12. If the mean hour is known, and it is only a question of 
maintaining it, the apparent motion of the fixed stars is easily used 
for this purpose ; in short, since these stars return to the same 
position at the end of 24 sidereal hours, it is sufficient to place a 
level of any sort, but fixed and invariable, in the direction of a star ; 
the next day at the same hour (solar time), less 3 minutes, 55.91 
seconds this same star would present itself anew before the level. 
The fixed stars afford the greatest facility for the control of the 
running of watches and clocks. 

13. Determination of the position of a point on the terrestrial 
Sphere. Since marine chronometers are among the instruments 
which are used to determine the position of any point on the surface 
of the earth, especially that of a vessel at sea, each watchmaker 
should inform himself of the part that these instruments play in 
such observations, on which depend the security of the ship and 
that of the beings which it transports. 

14. In order to represent the position of a point on the surface 
of a sphere, such as the earth, we suppose described on this 
sphere two great circles, one, passing through the two poles, 


General Principles of Cosmography. 


15 


is called the meridian circle; the other, perpendicular to the 
first, is the equator. This last is consequently perpendicular 
to the terrestrial axis, and at all points equally distant from the 
poles. Each of these circles is divided into 360 degrees. The 
divisions marked on the meridian circle commence at the equator, 
and are reckoned north and south to the poles, therefore from 
o to 90 degrees. These degrees are called degrees 0f latitude. 

Since we can imagine an infinite number of meridians passing 
through the poles, the point 0 can be placed arbitrarily, that is to 
say, the first meridian at whatever place it suits best. Thus 
England has chosen as the starting point the meridian which passes 
through the Observatory at Greenwich, in the neighborhood of 
London ; France has made choice of the one which passes through 
the Observatory of Paris, and other nations have made their first 
meridian pass through the Isle of Fer. 

The degrees reckoned on the equator are called degrees of 
longitude , and are reckoned both to the east and to the west of the 
first meridian, from o to 180 degrees. 

By imagining circles parallel to the equator passing through 
each division of the meridian circle, and meridian circles passing 
through each division of the circle of the equator, the latitude of a 
point will be then the distance in degrees from the parallel circle 
passing through this place to the equator, and its longitude will be 
the distance in degress from the meridian of this place to the 
meridian chosen as the starting point. 

These values constitute what are called the geographical 
coordinates of a point, and the position of this point on the terres¬ 
trial globe will be perfectly determined when we know its longitude 
east or west, and its latitude north or south. Thus we would say 
that the geographical coordinates of the city of Neuchatel, in 
Switzerland, are 

46° 59' 15 " north latitude, 

4° 35 7 54 " longitude, east of the meridian of Paris. 

15 . In order to determine practically the latitude of a point A, 
Fig. 3, the simplest manner is to measure the angle formed by a 
horizontal line A B , and by the line A C ending in the Polar star. 
In short, the fixed stars being prodigiously removed from the 
earth, which is but a point in relation to this enormous distance, 
we can say without appreciable error that all the straight lines 


i6 


Lessons in Horology. 


drawn from the earth to the Polar star are parallel to each other, 
which moreover conforms to experience. 

Since the Polar star is found almost exactly on the prolongation 
of the axis of the earth, the straight line that we imagine drawn 
from any point on the globe to this star is parallel to the axis. 
We can then say that the latitude of such a place as A , which is in 
reality the angle A O E } is represented by the angle B A C; 



practically the angles A O E, and B A C are equal by having their 
sides perpendicular each to each. We see then that the latitude of 
a place is equal to the height of the pole above the horizon. In 
order to determine it with greater exactness it would still be neces¬ 
sary to take note of the distance of the Polar star from the axis and 
of the refraction of the luminous rays. 

16 . The means employed in order to determine the longitude 
by direct observations are scarcely practicable, therefore we use in 
preference a marine chronometer whose daily rate is known. 

Let us suppose that a ship is one day at a certain point, which 
we will designate by A, that we have made the observation of the 








General Principles of Cosmography. 17 

passage of the sun across the meridian, and that we have noted the 
advance or the delay of the hour marked by the chronometer at 
this instant. Let us call C Q the number of hours, minutes and 
seconds shown by the watch at the exact moment of the passage of 
the sun across the meridian. One or several days later, we repeat 
this observation and we find, let us suppose, a new correction that 
we will designate by C. 

The difference between C 0 and C t divided by four minutes, 
will give us the number of degrees that the ship has advanced in 
one or the other direction of longitude. 

In short, since the earth takes twenty-four hours to accomplish 
a complete rotation of 360 degrees, it would require which is 
four minutes to cover one degree. We understand that it would 
still be necessary to take account in the calculation of the daily rate 
of the chronometer ; this value is an average based on a great 
number of former observations. All the accuracy of this method 
depends then on the absolute regularity of the daily rate of the 
marine timepieces. 

The values C 0 and C are algebraic values, that is to say of 
quantities preceded by positive signs (-j-) or negative signs (—), 
since the chronometer can be behind or in advance at the time of 
the passage of the sun across the meridian. 

Let us determine now which of these signs belongs to the 
advance of the timepiece and which will designate its retardation. 

In the first place, it is evident that this choice is arbitrary ; 
thus, at the Observatory of Greenwich, they have adopted the 
negative sign for the retardation and the positive sign for the 
advance. This choice seems at first sufficiently normal, but at 
Neuchatel one is placed at a different point of view. When the 
chronometer is slow at the passage of the sun, in order to find the 
true time we must add to the hour indicated by it the correction 
C ; this value should then be preceded by the positive sign, while 
when the chronometer is fast at the passage of the sun we must 
subtract the correction C in order to obtain the correct time ; in 
this case then,' this value should receive the negative sign. It is 
this manner of viewing the matter which has led, at the Observatory 
of Neuchatel, to the adoption of the sign -f for slow, and — for 
the advance of the chronometer 

IT. As we have said that a chronometer never follows exactly 
the mean time, its daily rate therefore should be determined in an 



i8 


Lessons in Horology. 


observatory ; because this daily rate must be reckoned with in the 
determination of the longitudes. Let us designate by A this mean 
value, and since it can be fast or slow let us precede it by the 
negative or positive sign. The chronometer is observed at the 
moment of the passage of the sun across the meridian ; this obser¬ 
vation giving the true time it will be necessary to deduct the mean 
time from it. Let us call B the difference between the mean time 
and the true time (equation of time) and let us determine the sign 
of this last value. 

Since it is desired to bring back the correction C to the mean 
time we will argue that, if the mean time at noon is fast, the true 
time is slower than the mean time and the value B should receive the 
sign on the other hand, if the mean time is slower than the true 
time, B would receive the sign —. 

Let suppose now that a ship leaves a seaport whose longitude 
is E 0 degrees west of Paris. The day of departure we have ob¬ 
served the passage of the sun and obtained a correction C 0 . The 
correction C ' 0 of the chronometer on the mean time will be for this 
day 

C' o = C 0 T B 0 

at the end of ./V'days we repeat the observation of the passage of 
the sun, and we will obtain a correction C between the time of the 
chronometer and the true time ; the correction C' between the 
time of the chronometer and the mean time will be expressed by 

C' = C+ B. 

The difference D between the time of departure and the time 
of the place where the vessel now is will be 

D=(C+ B)-(C 0 + B 0 )-NA. 

Reducing this value to minutes, we will have the longitude E 
in degrees by the division : 

E— . (£"+ B ) ~ (<P 0 + B n ) — NA 

~4 + °. 

18 . Let us take a numerical example. 

The longitude of Havre west of the meridian of Paris being 
2 ° 1 3 ' 45 ”> l et us imagine a vessel leaving this port November 2, 
1:893. The time shown this day by a marine chronometer at the 
moment of the passage of the sun across the meridian is 11 h. 




General Principles of Mechanics. 19 

43 m. 42 s. The equation of time for this date is + 16 m. 21s. 
We will then have noted the correction C Q = 11 h. 43 m. 42 s. 
-f- 16 m. 21 s. = 12 h. o m. 3 s. Four days after, a new 
obseivation shows that at the moment of the passage of the sun 
the chronometer indicates nh. 28m. 57s. For the 6th of Novem¬ 
ber, the equation of time being -f 16m. 15s. the new correction 
will be C = 11 h. 28m. 57s. -f- 16m. 15s. = 11 h. 45m. 12s. 

The difference 


D — 12 h. o m. 3 s. — 11 h. 45 m. 12 s. — N A 
gives in subtracting 

D = 14 m. 51 s. — N A. 

Supposing that the mean daily rate of the chronometer be A 
= — 0.5 s. we will have 

NA = 4 X — 0.5 s. — — 2 s. 
then D — 14 m. 51 s. — 2 s. = 14 m. 49 s. 

we will have then the longitude sought by 

, 7 14 m. 49 s. , 

E = ~ 4 + 2 I 3 / 45" 

by performing the division we will obtain 

P = 3° 42' 15" + 2 0 if 45" = 5° 56'. 

The ship will then be at noon on the sixth of November 
longitude 5 0 56' west of Paris. 

II. General Principles of Mechanics. 

» 

19. Forces. Any cause which produces or modifies the move¬ 
ment of a body is a force. A force can be power or resistance , that 
is to say, it can, without losing its active character, act in the same 
manner as or contrary to the movement. Such are the effects pro¬ 
duced by animated beings, by wind, steam, waterfalls, etc. 

Passive forces exist naturally and can partially or totally 
destroy motion, but are incapable of producing it; such are, among 
others, the effects produced by friction, the resistance of the air, etc. 

20. We can estimate very accurately the greatness of forces 
by their effects. The value of a force can always be represented 
by a weight, as the kilogramme or gramme, which would make 
equilibrium with it. Thus the force exerted by a man in order to 
put a car in motion can have its greatness measured by a certain 



20 


Lessons in Horology. 


number of kilogrammes. Let us suppose, for instance, a cord 
fastened to the car and passing over a fixed pulley placed before 
the vehicle ; if we suspend weights to the free end of the cord and 
increase them until the car commences to move, the total of the 
weight will give us the measure of the effort put forth by the man 
in order to produce the movement desired. 

21 . As a general rule, we give the name of motive force to any 
power which puts a body in motion, and, on the other hand, that 
of resistant force to every active or passive force in opposition to 
this movement. 

22. Without being able to define the nature of forces, the 
sensations which they invariably produce in us give us immediately 
an idea of their intensity and of their directio?i. 

The directions of forces are represented by the straight lines 
along which they tend to move the body to which they are applied. 
It is suitable to represent their intensity by lengths which are pro¬ 
portionate, the result is that we can submit forces to the same 
mathematical processes as any other quantities. 

23. The point Of application of a force is that part of a body 
on which it acts directly in order to change the state of motion or 
of rest of this body. 

24. The line Of direction of a force is that along which it tends 
to make its point of application advance. 

25. A force capable of replacing by itself alone a system of 
forces acting on a body is called the resultant of all these forces. 
These last are called the components of the only force able to 
replace them. 

26. The trajectory is a line which the movable point follows. 
The movement is called rectilmear or curvilinear , according as the 
trajectory is a straight line or a curve. 

27. Law of inertia. Experience has established a law to 
which all bodies are subject and which constitutes a fundamental 
principle of mechanics. This law known under the name of “ prin¬ 
ciple of hiertia ’ ’ can be defined as follows : 

A material body cannot put itself in motion if it is at rest and, 
reciprocally , if it is in motiori it caniiot of itself modify its movement. 

28. Definition Of mechanics. Mechanics is the science of forces 
and their effects. Its object is to find the relations of the forces 
which affect a body or a system of bodies causing this body or this 
system to take a certain movement in space. Reciprocally being 


General Principles of Mechanics. 


21 


given a body or a system of bodies, to find the motion that this 
body or system of bodies will take in space under the action of 
given forces. 

This general problem comprehends the one in which the forces 
make no change in the state of the body or of the system, a 
particular case in which we say that the forces are in equilibrium. 
Thence comes the division of mechanics into statics or the science 
of equilibrium and dyyiamics or the science of motion. We can 
still study the movements of bodies, considering only their direc¬ 
tion, intensity and duration, in leaving out the matter of which the 
bodies are formed and the forces which produce or modify these 
movements. This study forms a part of mechanics to which is given 
the name of kinematics, which can also be called geometric mechanics. 

29. Motion. Motion is uniform , when equal distances are 
traversed in equal times. 

We call the space traversed in the unit of time velocity , we 
will have then, designating velocity by v, space by s and time by /, 

s = t v 

whence we have 

v = q and t = \ 

30. The motion is called variable if the spaces traversed in 
any equal times are unequal ; that is to say, when the speed of the 
body is not constant during the entire duration of the motion. 

31 . When a moving body traverses certain equal distances in 
equal times, without fulfilling the same conditions for parts of these 
distances, we say that the motion is periodically tcniform. Such 
are, for example, the motion of the earth around the sun and the 
vibratory motion of a pendulum in small amplitudes. 

32. The motion is uniformly variable when the velocity of the 
moving body varies equal quantities in equal times. 

The acceleration is then the quantity which the velocity varies 
during the unit of time. 

If in uniformly variable motion, the velocity increases the 
acceleration is positive and we say that the motion is uniformly 
accelerated. 

If the velocity diminishes, the acceleration is then negative 
and the motion is said to be uniformly retarded. 

33. The motion of a body which falls by the action of its 
weight is uniformly accelerated. In this case we designate the 
acceleration due to the weight by the letter g; this value is constant 


22 


Lessons in Horology. 


for the same place and in our regions g = 9.8088 m. This value 
represents twice the distance traversed during the first second by 
a body falling freely and without initial velocity. 

34. Rotary motion. A solid is animated with a movement of 
rotation on an axis when each of its points describes a circum¬ 
ference whose plane is perpendicular to the axis and whose center 
is found on this axis. In this movement any two points of the body, 
describe, in the same time, similar arcs, that is to say, of the same 
number of degrees ; but the lengths of these arcs are different 
and should be proportionate to their distances from the axis. Let 
e and d be the arcs traversed in the same time by two points m 
and m' (Fig. 4) situated at the distances rand r' from the axis of 
rotation, we would have 

e r 

e' r' 

The movement of rotation is uniform if, in equal times, a point 
of the body describes always equal arcs. 

The velocity of such a motion 
can only be determined by con¬ 
sidering at the same time the 
path traversed in 1 second by a 
point of the body and the distance 
of this point from the axis of rota¬ 
tion. In order to avoid this double 
data, we consider the points which 
are at the unit of distance from 
the axis and we call the angular 
velocity the length of the arc de¬ 
scribed in 1 second by a point 
situated at the unit of distance 
from the axis. 

Let w be this arc, we will have, for the velocity v of another 
point situated at the distance r 

v — r w, 

from whence we conclude w = L Consequently, the angular 
velocity is obtained by dividing the space which is traversed by 
any point in 1 second, by the radius of the circumference which 
it describes. 

35. Mass Of a body. The quotient of the weight of a body 
in any place on the globe, by the acceleration due to gravity at this 




General Principles of Mecka?iics. 


23 


place is constant. This value is what is called in mechanics the 
mass of the body. This quantity, which is of a particular nature 
since it is nothing else than a quotient, can be subjected to calcula¬ 
tion just as any other quantity. In designating it by M, we have 

M = — from whence M p- = W and g = ~ 

36. The product M v of the mass M of a moving body by the 
velocity which it possesses, takes the name of quantity of motion. 

Work of a Force. 

37. Definition. We call in mechanics work of a force the pro¬ 
duct of the intensity of this force by the path traversed by its point 
of application. In other words : 

The work produced by a force constant in magnitude and in 
direction is represented by the product of the intensity of this force 
by the projection of the distance traversed by the point of applica¬ 
tion on the direction of the force. 

Thus, the path and the force being in the same direction, we 
will have (Fig. 5) 

W = F X A B. 

If the path A B and the force F have not the same direction, 


A B 2 

Fig. 5 

we will project in this case the path A B on the direction A F and 
we will have (Fig. 6) 

W =FXAC 

Let us observe that the projection of the path on the force is 
greater as the angle B A C becomes smaller ; the work will then 



be greatest when A B and A F will have the same direction. If 
the angle B A F becomes greater, the projection diminishes, and 




24 Lessons in Horology. 

becomes nothing when the angle is equal to 90°; in this last case 
the work is null. 

38 . Work of a force tangent to a wheel. On imagining the 
movement of rotation of the wheel to be very small, we can admit 
that the movement takes place along the tangent. Let us call 
F the force and s the space traversed in its direction, we will have 
for this slight displacement 

W = FX s. 

When the wheel will have made a complete revolution, the 
path s will have become a circumference and we will have the work 
for one revolution expressed by 

IV = F X 2 7T r. 

39 . Unit Of work. We have chosen for unit of work that 
which is the product of the unit of force by the unit of distance, 
that is to say, in mechanics, of the kilogramme by the meter and 
in horology, of the gramme by the millimeter. We have given to 
this unit the name of kilogrammeter for the machines and of 
grammillimeter for the more delicate pieces of horology. As 
abbreviation, we will designate by the letters gr. m. this last unit 
which we will make use of in the entire extent of this course. If, 
for example, the force is 3.5 grammes and the distance 0.4 milli¬ 
meters, the work of the force will be 

W — 3-5 X 0.4 = 1.4 gr. m. 

40. Active power. A weight P which falls from a height h 
generates a certain work that we can represent by the product 

FX k. 

We find in mechanics that every body falling from a height h 
is animated by a velocity v which is connected with the height h by 

the relation v 2 = 2 gh, 

from which expression we can draw 



Replacing in the equation of work Ph , h by this last value, we will 
have 

which we can also write 



but ~ being the mass m of the body, we will have at length 

Ph — \ m v 2 . 




Ge?ieral Principles of Mechanics. 


25 


The expression \ m v 2 has received the name of active poiver. 
We can then say that the active power of a body in motion is half 
the product of its mass by the square of its velocity ; or, also, that 
the mechanical work which imparts a certain velocity to a body is 
equal to the active power which animates that body. 

We give the name of active force to twice the active power ; 
we have then 

Active power = \ m v 2 , 

Active force = m v 2 . 

41. Every body in motion is capable of doing work. In effect 
the body has a mass m ; it has a velocity v, since it is in motion, con¬ 
sequently the product \ m v 2 gives us the value of the work Ph , 
to which the velocity v corresponds ; we can therefore say that 
every body in motion is capable of producing work. 


Moment of a Force. 


42. Let us now imagine two cylinders of different diameters 
turning around an axis O (Fig. 7) and let us admit, for example, 
that the first is three times as great as the second. Let us 

suspend weights at the ends of light cords 
wrapped around each cylinder, in such a 
manner that each of these weights acts in a 
contrary sense to the other. In order that 
equilibrium may exist in this system, we will 
find that the weight fixed to the small cyl¬ 
inder should be three times as great as 
that which is fixed to the large one. In 
this manner, if we turn the cylinder one 
revolution, one of the weights will rise 
while the other will fall ; the weight p tra¬ 
versing a path represented by 2 - X 3 its 
work will be 

2 ft X 3 A 



The weight P traverses a path 2 iz X 1, producing at the same 
time a work 

2 t x 1 P, 


Since we have 3 p — P we can admit 

(a) 2 7 T X 3 P = 2 ir X 1 P, 

and there will be equilibrium, because the mechanical work of one 
of the weights is equal to that of the other. The equality of the 





26 


Lessons in Horology . 


work of these two weights will exist also when we make the cylin¬ 
ders describe only a fraction of a revolution : as small as this 
fraction may be. 

Dividing the equation (a) by 2 7r we obtain 

3 p — i p. 

The figures 3 and 1 are the respective lengths of the radii of each 
cylinder ; this radius takes the name of lever arm and the above 
product of the intensity of the force by the lever arm is called the 
moment of the force. 

Summing up, we have just examined the state of a body 
which can turn around a fixed point ; to this body are applied two 
forces whose work mutually counteracting, produces equilibrium. 
We call such a system a lever. 

In every lever, for equilibrium to exist it is therefore necessary 
for the moments of the two forces in action to be equal. 

Imagining the system in motion, under the action of an exterior 
impulse, we will find the work of the forces on multiplying their 
moment by the angle traversed ; in this new condition of the body 
the work of the forces will then be equal. 

43. Lever. Practically, a lever is a solid body, movable 
around a fixed point and acted upon by two forces tending to 



make it turn in contrary directions. Fig. 8 represents a lever 
in which O is the fixed point. P and F the two forces. The 
lever arms of the forces P and F are the distances from the fixed 
point O to the two forces, that is to say, the perpendicular O A f 
and 0 B’ dropped on the direction of these forces. 



General Principles of Mechanics. 


27 


From what we have said before, equilibrium will exist when 
the moment of the force P will be equal to the moment of the 
force F, that is to say, when 

FX 0 A' = PX 0 B\ 

which can be written 

JP_0 B' 

P~ O A' 

The two forces should, therefore, be in inverse proportion to their 
lever arms. 



44. One often distinguishes two kinds of levers : In the lever 
of the first kind the fulcrum is situated between the points of 
application of the two forces ; in the other, Fig. 9, this fixed point 
is situated at one of the extremeties of the body. From the 
theoretical point of view this distinction is useless and the con¬ 
ditions of equilibrium of the lever apply to all cases. 

Transmission of Work in Machines 

45. We give the name of machine to every system of bodies 
intended to transmit the work of forces. In order to explain in 
what manner this transmission is effected it is necessary to enter 
into some details. 

The relative movements of the different parts of a machine are 
not determined only in direction but also in intensity. Generally 
the movements are periodically uniform (31) ; the speed is put in 
harmony with the requirements of the industrial work to be 
produced without its ever attaining the limit at which the solidity 
of the machine would be endangered. 

46. Different forces act on a machine in motion, which can be 
divided into three classes : 

1st. Motive forces. These are those which act in the direction 



28 


Lessons in Horology. 


of the movement of the parts which they operate ; it is consequently 
to these that is due the motion of the machine. 

2d. Useful resisting forces, those which the materials on which 
the machine operates, oppose to the movement of the parts which 
act on them ; it is these then which we desire to overcome. 

3d. Passive or hurtful resisting forces, which arise from the 
movement of the different parts of the machine to oppose this 
movement. We have already seen that they are due to the friction 
of these parts among themselves or on foreign bodies, to shocks 
which can be produced between these parts on account of sudden 
changes in speed and to the resistance of the air. 

Considering the motive forces as positive, since they act in the 
direction of the movement, the useful resistances and passive 
resistances will then be negative. If we suppose the system 
animated by a uniform movement, the algebraic sum of the work of 
all the forces for any given time will be null, since the gain or the 
loss of active power is null, and we will have, in designating by 
W m the work of the motive forces, W u the useful work and W p the 
work of the passive forces : 

Wm — Wu — W P = O, 

from whence 

Wm = Wu + Wp, 

which shows us that, the movement being uniform, the motive work 
is equal to the useful work, augmented by the work of the passive 
forces. 

When in any machine this formula is verified, we say then that 
there is “dynamic equilibrium.” 

When the movement of a machine is periodically uniform, the 
gain or the loss of active power is null only for a whole number of 
periods ; for this time we still have 

Wm — IVu “b W^p, 

We say then that the machine is in “periodic dynamic equi¬ 
librium;” this is the ordinary state of machines, not only on 
account of the shape of their parts, but because of the variations 
more or less great in the motive forces, and especially in the 
resistances. 

Thus W u is always inferior to W m ; that is to say, a machine 
renders less useful work than the motive power applied, because 
the work of the passive resistances is never null. 


General Principles of Mechayiics. 


29 


4 T, Calling P the motive force acting on any machine, and Q 
the useful resistance overcome by this machine, E and e being the 
spaces traversed by the points of application of P and of Q in the 
direction of these forces and in any equal time, at the beginning 
and at the end of which the speed of the machine is the same, the 
equation of dynamic equilibrium gives, supposing first that the pas¬ 
sive resistances are null : 

PE^= Qe, or ^ | 

From the equality between the work and the power and that 
of the resistance, it follows that for the same motive work P E , 
according as the force Q may be multiplied by j 4 , 2, 3, etc., 
the space e will be divided respectively by the same numbers ; from 
whence comes the maxim well known in mechanics: “That which 
we gain in force we lose in speed, or what amounts to the same 
thing, in distance, and reciprocally.” 

The preceding proportion enables us to calculate any one of 
the four quantities P f Q , E , e, when we know the three others. 

For any simple or complicated machine, if the question is to 
find the resistance Q that a power P can overcome, we determine 
the spaces E and e traversed in the same time by the points of 
application of the forces P and Q. E and e are any distance 
whatever if these points of application have uniform movements, 
but we take them corresponding to a period if the movement of the 
machine is periodic. When the machine is constructed, by putting 
it in motion of any sort, we determine the values of E ; we deduce 
those of e from the relations of the spaces traversed by the different 
parts which transmit the movement of the point of application from 
P to that of Q. 

Let us suppose that the resistance to be overcome Q be 100 
kilogrammes, and that it is desired to determine what will be the 
power of P, neglecting the passive resistances. We commence by 
determining the corresponding values of E and e f as has just been 
shown. Let E= 2.5 m., and ^ = 0.80 m.; replacing the letters by 
their values, in the preceding equation we will have 

P 0.80 

100 k 2.5 

from whence 

„ 100 X 0.80 , 

P= —h- 5— 32 kilogrammes. 

z, 0 

If we had known P we would have been able to determine Q, as 
we have just done for P. 




30 


Lesso?is in Horology. 


48 . In machines, especially in industrial machines, the passive 
resistances are so considerable that we cannot neglect the work that 
they absorb ; the dynamic equilibrium is then expressed by 

w m = w u + W p . 

For a certain displacement of the parts of the machine, the 
work W m W u W p will be valued as in the preceding case ; thus 
P being the power, Q the useful resistance, R , R ', .... the different 
passive resistances, and E, e , i, i’ ... . the corresponding distances 
traversed in the same time by the points of application in the 
direction of these forces, we have 

PE — Q e + R i + R' i' -f_ 

49 . It may happen that one or several hurtful resistances come 
from the shocks between the parts of the machine. The work 
absorbed by these resistances is no longer valued by the product 
of a force by the distance that its point of application traverses, but 
by the loss of active power due to the shock, and this loss, valued 
in units of work, enters into the second member of the equation as 
the other hurtful works R i R’ i’... . 

By the aid of the preceding equation, knowing in a machine 
two of the three following works : the W m — PE , the W u —Q e 
and the W p = R i R' /' + ... . we determine the third. 

50 . Ordinarily one decides to set up a machine capable of 
producing a given useful work. 

Wu — Q e 

It is then necessary to determine the lV m —P E , capable of 
producing not only this useful work but of overcoming also the 
secondary resisting works. 

One should then commence by calculating this hurtful work, 
which is done by determining the values of the different passive 
resistances R, R' ... . in function of Q, and afterward W p in 
function of W u . 

Having W p and W u , we can determine the value of W m 
expressed as has been said in kilogrammeters and in grammilli- 
meters. 

51 . The motive work W m being represented by ioo, the useful 
and hurtful works W u and IV P being for example, 75 and 25, the 
loss is then 25 for 100; we say in this case that the product of the 
machine is 75 per cent. If it were possible for the loss to be 
nothing, the product would be 100 per cent ; this fact can never be 



General Principles of Mechanics. 


3 i 


realized, which renders absolutely illusive the hypothesis of perpetual 
motion. The product of a machine rarely passes 80 per cent.; it 
is nearly always much inferior to this limit. 

In this preliminary study we have desired to establish a basis 
which is nothing more than the enunciation of some fundamental 
principles of mechanics. In the course which is about to follow, we 
will make a constant use of them, and all their developments will 
be found in the text. 


CHAPTER I. 


General Functions of Clocks and Watches. 


The Oscillations of the Pendulum and their Relation to the Motive Force. 

52 . We know that in clocks and watches time is measured by 
the periodically-uniform movement of the pendulum or of the 
balance wheel. 

History relates that Galileo, while yet young, was struck with 
the regularity of the pendulous vibrations of a candelabra in the 
Cathedral of Pisa. He studied the laws of these oscillations and 
used a pendulum later on for his astronomical observations. This 
instrument, in its primitive simplicity, presented two difficulties ; 
when the astronomer left his pendulum to itself, after having 
diverted it from the vertical position, the oscillations which were 
produced having at first a certain amplitude, diminished little by 
little, then finally ceased entirely. He was then obliged, from time 
to time, to give an impulse to his pendulum. The second of these 
difficulties was the necessity for him to count the number of these 
oscillations. It is said that he charged a servant with the execution 
of these two functions. 

Now, the mechanism of the clock performs unaided these two 
functions with a regularity that the man could never achieve directly. 

53 . Let us seek, in the first place, for the causes which make 
the oscillations of a free pendulum constantly diminish. 

When a pendulum is moved from its position of equilibrium 
O A (Fig. io), the attraction of the earth, which was perfectly 
neutralized by the resistance of the point of suspension O, is no 
longer so in the oblique position O B. It would cause the ball to 
descend vertically if the cord did not force it to describe the 
arc of a circle; at each instant of its returning course the speed 
of the pendulum increases a small quantity until it reaches anew 
the vertical position O A. From there on the inertia, or, if one 
prefers it, the velocity acquired, forces it to continue in its motion 
and makes it describe the arc A B '; from that instant also gravity, 
acting in the contrary direction of the motion, tends to stop it. 
The velocity diminishes constantly, and would become null at 
the moment when the ball would arrive at a height equal and 


32 




General F 7 cnctio?is of Clocks a?id Watches. 


33 


symmetrical to that from whence it started, if there were no other 
forces than those of gravity which would act on the pendulum. 

These forces which exert their action in the contrary direction 
to the motion, are resistances of the suspension and of the air; it is 
then to these that is due the diminution of the amplitude of the 
pendulum’s oscillations. If these forces could be suppressed, the 
motion would be perpetual. 

54 . There are two O 

manners of suspending a 
pendulum—by means of a 
knife edge and by means of 
flexible springs. 

The knife-edge siis- 
pension is made in such a 
manner that the friction is 
very slight, without its be¬ 
ing, however, completely 
annulled. This kind of 
suspension can be used in 
regulators whose amplitude 
of oscillation is generally 
small. For this purpose a 
sort of knife blade slightly 
rounded on its edge, and 
working in the interior of 
a hollow cylinder, is fixed 
transversely to the pendulum rod (Fig. n). The knife edge, as 
also the hollow cylinder, should be made of exceedingly-hard 
material and thoroughly polished. The knife edge can then be 
regarded as a pivot of very small dimensions. (We will see later 
on that the work of friction is proportional to the pressure and 
to the greatness of the amplitude.) 

The sprmg suspe?ision consists in terminating the upper end 
of the pendulum rod by two short blades of steel securely fastened 
on the other end to any fixed piece. In this system, which is also 
very much used, there exists a loss of force resulting from the 
distorting of the blades. 

55 . Concerning the resistance of the air , we admit that it is in 
direct relation to the largest transverse section of the body and to 
the square of the velocity with which it traverses the atmosphere ; 








34 


Lessons in Horology. 


it depends, moreover, whether its form is more or less tapering. 
The work of this force should be proportional to the cube of the 
size of the amplitudes. 

56 . Now that we know the nature of the forces which act on 
the pendulum during its vibrating motion, u'e can determine their 
work and establish the relation which connects them with each other. 

The motive 7 vork W m developed by gravity during the 
descending half oscillation, is equal to the weight P of the pen¬ 
dulum multiplied by the projection of the arc B A (Fig. io) on 
the direction O A of the force ; therefore, by the length Ab (37). 
We will write then 

Wm — P A Ab 

The resisting work , that is to say the work of the forces which 
act in the contrary direction to the motion, is composed of two 

distinct forces : 

1 st. Of the work of gravity developed while 
the pendulum traversed the half oscillation ascend¬ 
ing, therefore the weight P multiplied by the pro¬ 
jection of the arc traversed A B' on the direction 
O A ; let us represent this work by the formula 

Wu = P X Aa 

2d. Of the secondary resisting works arising from the resist¬ 
ances of the suspension and of the air. Knowing the lengths of 
the arcs A B and A B ', we find the work of the secondary resisting 
forces by multiplying the weight P of the pendulum by the 
difierence of the projections Ab — Aa or ab; we will then have 

Wp — PA ab 

The motive work should be equal to the sum of the resisting 
works (46) ; we will therefore have 

Wm = Wu+ W P , 

or substituting 

P A Ab = P A Aa + P A ab 

57. For the oscillations of the pendulum to preserve the same 
amplitude, it is therefore necessary that at each of these oscillations 
it must receive an impulse whose work should be equal to P A ab. 

58 . Since the secondary resisting work increases with the 
amplitude of the oscillations, it is necessary that the impulse, or 
what we should call the work of the maintaining force, should be 
greatest w r hen we wish the pendulum to traverse the largest arcs. 






General Functions of Clocks and Watches. 


35 


We see also that the more we diminish the friction of the 
knife edge and the resistance of the air, the less maintaining force 
is necessary. We diminish the resistance of the air by using a 
pendulum ball of high specific gravity, because for such a weight 
the section which traverses the air is smaller. The pendulum can 
also be suspended under a glass from which the air has been 
exhausted. 

59 . In order to maintain the oscillations of the pendulum in 
clocks we use most frequently motive forces produced by a weight, 
a coiled spring or an electric current. The two first will be the 
subject of a detailed study in the following chapter. 

The Oscillations of the Balance and their Relation to the Motive Force 

60 . The motion of the pendulum cannot be employed for 
measuring time, except in instruments which can maintain a fixed 
position. In portable timepieces we utilize the vibratory motion 
of an annular balance, mounted on an axis and furnished with a 
spiral spring. 

61 . This spiral spring is a thin blade of metal, of sufficient 
length, wound on itself in the form of the spiral of Archimedes, or 
of a cylindrical, spherical or conical helix. In each case, one of 
the extremities of this blade is fastened to the balance wheel and 
the other to a piece fastened to some part of the watch. 

When we place in a watch movement the balance wheel fitted 
only with its spiral, there is found a position in which the elastic 
force of the spring exercises no influence on the balance. The 
latter is then in the condition of repose. 

When we move the balance from this position in either direc¬ 
tion the elastic force of the spring tends to bring it back to the 
point of repose ; there are then produced oscillations analagous to 
those of the pendulum. 

This oscillatory motion is very useful for measuring time and 
has the advantage of being suitable for employment in all portable 
timepieces. 

62 . Suppose A (Fig. 12) the point of repose of a balance 
wheel ; if the latter be moved from that position the angle 
A O B = a, and if at the point B it be released, thus allowing the 
elastic force of the spring to act on it, this force will impart to it a 
movement of rotation whose speed will increase up to the point A. 
Passing that point, the spiral will exert a force contrary to the 


36 


Lessons in Horology. 


direction of the motion and tending to stop it. If it were possible 
to produce such a movement without there being any passive 
resistances acting on the balance wheel, the latter would traverse a 

new angle, A O b — a, 
then would come back to B, 
and so on indefinitely. 

It is not so in reality, 
for there are a number of 
resisting forces which act on 
the balance and which pre¬ 
vent it from arriving at b. 
These forces are : 

1. The friction of the 
balance pivots. 

2. The resistance of the 
air. 

3. A loss of force resi¬ 
dent in the spiral, the true 
cause of which is not abso¬ 
lutely defined but the existence of which can be perfectly established. 

These secondary resisting forces have the effect of diminishing 
each oscillation a small quantity, which is represented in the figure 
by the angle B' O b. Calling a ' the angle B’ O A , we have 

B f O b = a — a'. 





A 

Fig. 12 


If, as we have done for the oscillations of the pendulum, we 
designate by W m the motive work exerted by the spiral while the 
balance wheel traversed the angle a, W u the resisting work pro¬ 
ceeding from the spiral during the second part of the oscillation ; 
therefore, while the balance wheel traverses the angle a', and W p 
the secondary resisting work of the passive forces, we would 
obtain the equality (46) 

Wm = U'u T Wp , 
or 

Wp = Wm — Wu . 

The work of the maintaining force should be, both for the 
balance and for the pendulum, equal to the secondary resisting 
work, if you wish to preserve the initial greatness of the amplitude 
of the oscillations ; otherwise expressed, the work of the maintaining 
force should be equal to the work of the force of the spiral while 
the balance wheel traverses the angle a — a'. 








General Functions of Clocks and Watches . 


37 


63 . We can admit that the resisting work increases with the 
amplitude of the oscillations, as we have shown for the pendulum, 
and conclude that more motive work would be necessary to tra¬ 
verse larger arcs than for smaller ones. 

64 . We use exclusively for motive force in portable timepieces 
the elastic force developed by a spring enclosed in the interior of a 



Fig. 13 


cylinder called the barrel . This piece, generally toothed, turns 
around an axis, and this action is conveyed to the balance wheel by 
special mechanism, which we are going to pass rapidly in review. 

Wheel-Work. 

Its Purpose in the Mechanism of Clocks and Watches. 

65 . The motive force, not acting directly either on the pendu¬ 
lum or on the balance wheel, is first transmitted by a system of 
toothed wheels or train of gearings that is called in technical lan¬ 
guage the wheel-work or the transmission. This force, thus trans¬ 
ported, is received by a mechanism which is the escape?ne 7 it ; it is 







38 


Lessons in Horology. 


this last whose function it is to restore to each oscillation of the 
pendulum or balance whee 1 the loss of force, 

Wm — IVu , 

occasioned by the secondary resisting forces. 

66 . When a weight is used as motive force, that weight is sus¬ 
pended to the extremity of a cord unwinding from a cylinder fixed 
concentrically on the axis of a toothed wheel. This wheel A 
(Fig. 13) gears in a second wheel much smaller than the first and 
which is called a pinion , on which is fixed concentrically a second 
toothed wheel B , which in its turn gears in the pinion b , and so on 
to the last pinion, on whose axis the escape wheel is fastened. The 
same thing takes place when the motive force is that of the spring 
in the barrel. In this case the barrel gears directly into the first 
pinion a. 

67 . The different wheels of the wheel-work in watches bear 
the following names : 

1. The barrel. 

2. The center wheel * (large intermediate wheel). 

3. The third wheel (small intermediate wheel). 

4. The fourth wheel (seconds wheel). 

5. The escape wheel (escape wheel). 

The pinions carrying the four last mobiles take the same name 
as the wheel to which they are riveted. 

68. The mechanical work of the motive force is then trans¬ 
mitted by the wheel-work to the escape wheel. This transmission 
cannot be effected, however, in a complete manner, because part 
of the force is absorbed by the friction of the gearings and of the 
pivots, by the inertia of the moving bodies and sometimes also by 
the defects resident in the gearings. 

69 . Beside the transmission of the force, the wheel-work should 
fulfil another function : this is to reckon the number of oscillations 
that the pendulum or the balance wheel executes during a deter¬ 
mined time and to indicate this number by means of hands on a 
properly-divided dial. We must, therefore, combine the relation 
of the numbers of teeth in the wheels to the numbers of leaves in 

*We give here the appellations still in use in most of the horological workshops of the 
canton of Neuchatel. The names of large and of small intermediate wheels are really 
obsolete and should be replaced by the following, which are in better relation to the positions 
of these two mobiles : “ Center wheel ” for the first and “ intermediate wheel ” for the second. 
To avoid confusion, we conserve, however, in this edition the denominations in use in this 
locality. (The usual names employed in America have been made use of by the translator.) 



General Functions of Clocks and Watches. 


39 


the pinions, so as to make this indication conform to the division of 
time. Thus, the center wheel carrying on its axis the minute hand 
should complete one rotation during one hour, and the fourth 
wheel carrying the second hand should make one revolution each 
minute. 

(The hour hand is carried by a wheel forming part of an 
accessory wheel-work, which will occupy our attention later.) 

Escapements. 

70. Several kinds of escapements have been constructed, 
differing more or less from each other, but whatever they may be 
their function consists always in restoring to the pendulum or to 
the balance wheel the speed which the passive resistances have 
made them lose. The most perfect escapement will be the one 
which will effect this work by altering as little as possible the dura¬ 
tion of the oscillation. 

Since the movement of the balance wheel as well as that of the 
pendulum is an oscillating movement, the escape wheel is arrested 
during part of the oscillation ; it is only when the balance or the 
pendulum has traversed a determined arc that the wheel becomes 
free and is put in motion. During this time it acts either directly 
on the balance, as in the “ cylinder ” escapements or the “ detent,” 
or on an intermediate piece, as in the “anchor” escapements. 
After having traversed the angle of impulse determined, the wheel 
is arrested anew until another disengagement. The manner in 
which this arresting is produced differs according to the kind of 
escapement. 

71. In most of the escapements the action of each tooth of the 
wheel corresponds to two oscillations of the balance wheel or pen¬ 
dulum. Thus, in a watch, the balance wheel executes 30 oscilla¬ 
tions during one complete revolution of a wheel of 15 teeth ; in a 
clock, the pendulum makes 60 oscillations during one revolution of 
an escape wheel of 30 teeth. 

72. To recapitulate, the study of the functions of horological 
mechanism can be divided into four principal parts, which are : 

1. Pow r er—study of motive powers. 

2. Transmission—study of wheel-works and gearings. 

3. Reception—study of the escapements. 

4. Regulation—study of regulating and adjusting. 


40 


CHAPTER II. 


Maintaining or Motive Forces. 


The Weight as a Motive Force. 

T 3 . We will adopt in the beginning as units in the calculations, 
the millimeter as unit of length, the gramme as unit of weight and 
of force, which gives us for the unit of mechanical work the 
grammillimeter. We will choose the second as the unit of time. 

74 . Among all the forces which are used in horology in order 
to maintain the oscillations of the pendulum, the weight is at once 
the most regular, the most simple to obtain and the one whose 
intensity can be regulated with the greatest facility. 

75 . If a certain weight P (Fig. 13) is suspended at the end of 
a cord wrapped around a cylinder the radius of which increased by 
half the thickness of the cord is equal to r, the work of this force while 
the cylinder executes one revolution will be expressed by (38) 


P X 2 7 x r. 

Dividing this work by the number N of oscillations that the 
pendulum executes during one revolution of the cylinder, we will 
have as quotient the mechanical work developed by the weight 
during one oscillation of the pendulum, thus : 

W p — T' 

JV 

We know that a part of this mechanical work is lost during its 
transmission to the pendulum: calling W u this last work, we 
should have the equality 

W P — Wu -=-■ W P , 

in which we will replace W P by its value, thus : 


P x 2 r 

~N~ 


Wu = 


Wp . 


We see then that the determination of the work which the 
pendulum receives at each oscillation (IV P ) depends also on the 
knowledge of the work lost during its transmission by the wheel- 
work and the escapement. We understand, consequently, the 
difficulty that there is to determine the motive work, since this 
work does not depend alone on the weight and on the dimensions 
of the pendulum but also on the resistances to be overcome during 
an oscillation. 

Here are, however, two calculations taken from practice which 
will aid in more firmly fixing the ideas on this subject : 




Maintaining or Motive Forces. 


4 * 


76 . First Calculation .—The motive weight of a regulator 
beating seconds is 2000 grammes ; this weight is suspended at the 
end of a cord which unwinds from a cylinder, with a radius of 
15 millimeters. What will be the work produced by this weight 
during the unit of time? 

The mechanical work effected by the weight while the cylinder 
executes one revolution will be 


2000 X 2 ir X 15 — 188496 gr.m. 


A wheels is fastened to the cylinder (Fig. 13) gearing in a 
pinion which carries on its axis a second wheel B, which in turn 
gears into a pinion b, this last pinion carrying on its axis the minute 
hand should then execute one revolution an hour. The numbers 
of teeth and leaves of these moving bodies are distributed in such 
a manner that the pinion b executes 45 turns while the cylinder 
makes one ; consequently, one revolution of the cylinder takes 
place in 45 hours or in 

45 X 60 X 60 = 162000 seconds = N. 

We will then obtain the work produced by the weight during one 
oscillation of the pendulum, by the application of the formula, 


U P = 


P 2 ir r 

~~N~ 


188496 

162000 


1.163 gr.m. 


We will show the manner of calculating the work lost during 
the transmission when we treat of the questions of frictions, of the 
inertia of the wheels, etc.; for the present, let us admit these calcu¬ 
lations as made and adopt for this special case the value 


We will then have 
or 


IVu — 0.413 gr.m. 

Wm — IVu Up , 

1.163 — 0.413 = o -75 gr.m. 


The weight of 0.75 grammes, exerting its action on a dis¬ 
tance of one millimeter, is then sufficient to keep up the oscillations 
of a pendulum whose weight is about 6500 grammes. The ampli¬ 
tude of the oscillations is 2° 6'. 

77 . Although that which follows is a little outside of the problem which 
we have just solved, let us profit, however, by the data that we possess to 
calculate further the angle 

B O A — B' O A (Fig. 10). 

This adjunct to the preceding solution does not, moreover, lack in interest. 


42 


Lessons in Horology . 


Following an equation previously established (56), the work of the force 
capable of maintaining the oscillations of a pendulum was expressed by 

IVp = P X a b. 

We can then put 

P X a b — 0.75 gr.m.; 

or, again, 

* 0.75 

a b — -ggQjj- = 0.0001154. 

The length of a simple pendulum beating seconds is about 994 milli¬ 
meters for our latitude.* Let us suppose that the entire weight of our pen¬ 
dulum is assembled at a single point, the distance from the center of gravity 
to the center of suspension will then be equal to the length of a simple pen¬ 
dulum beating seconds. We will have 

A b — 994 — 994 cos A O B 
A a — 994 — 994 cos A O B'. 

From Fig. 10 the difference A b — A a gives the length a b ; sub¬ 
tracting then the two foregoing equations, one from the other, we obtain 

a b — 994 cos A O B' — 994 cos A O B, 

whence follows 

cosAOB'-cosAOB = %± = 

since the angle A O B is equal, in this case, to half of 2 0 6', which is i° g, 
we can write, after having completed the calculation of the second member 
of the equation : 

Cos A O B' — cos i° 3 / — 0.000000116. 

In order to determine the value of the angle A O B — AO B', we can find 
in a table of natural trigonometrical lines the difference between the cosines 
of the angles i°2 / and i° g. This difference is 0.0000053 ; we will then 
have the proportion, 

0.0000053 60" 

0.000000116 X 

from whence x — 
then A OB — A O B f = i.g'. 


78 . Seco?id Calculation .—A clock from the Black Forest, such 
as those that were manufactured in large quantities during the years 
between 1840 and 1850, runs under the action of a weight of 
625 grammes. This weight descends in 24 hours from a height of 
1250 millimeters. What is the work produced by this force during 
one second ? 

The work produced during the descent of the weight will be 

IV = 625 X 1250 


in 24 hours ; during one second it will be 24 X 60 X 60 = 86400 
times less; therefore 


ij? _ 625 X 1250 

H/m 86400 


9 gr.mm. 


* Latitude ol Neuchatel 







Maintaining or Motive Forces. 


43 


We see that this clock requires a much greater mechanical work 
than that of the regulator of the preceding example. This difference 
becomes still more obvious if we compare the two pendulums. 
The weight of the pendulum of the last clock is only 8 grammes, 
while the pendulum of the regulator weighs 6500 grammes. 

Although we could not, at this time, compare two clocks, 
whose pendulums have neither the same length, nor the same 
weight, nor the same amplitude of oscillation, we note, however, 
that the regulator requires much less motive force than a small 
clock of the Black Forest. 

The Barrel Spring as a Motive Force. 

T9. These springs are thin blades of properly-tempered steel ; 
they are of a sufficient length and coiled up in spiral form in the 
interior of the barrel. One of their extremities is fastened to the 
wall of the drum and the other to the hub , which is a cylindrical 
piece adjusted on the arbor of the barrel or forming part of it. 
When one holds firmly either the barrel arbor or the barrel, and 
causes the one of these two pieces left free to turn, the spring 
begins to wind around the hub and manifests a certain force from 
its extremities, which tends tb bring it back to its first form. When 
the arbor is made fast, the force displayed by the spring has then 
the effect of causing the barrel to revolve. 

80. The place occupied by the spring in the interior of the 
barrel should be equal to half the disposable space. 

81. Measurement of the Force of a Spring. The force developed 
by the spring is susceptible of measurement. For this purpose let 
us adjust on the barrel arbor a graduated lever arm, along which a 
certain determined weight can slide. While holding the barrel in 
the hand, let us set up the spring to the point that we wish to 
study, one turn for example ; let us endeavor then to produce 
equilibrium by sliding the weight along the lever arm. When the 
two actions, that of the weight on one side and of the spring on the 
other, neutralize each other, equilibrium is produced, and it is then 
evident that the effort displayed by the spring is equal to the 
effect produced by the weight. This last effect will be perfectly 
determined when we know the size of the weight and the length 
of the lever arm, at the extremity of which it exerts its action. 
We know that in mechanics the moment of a force (42) is the 
product of the intensity of this force by its lever arm. 


44 


Lessons in Horology. 


The moment of the force of the weight will give us then the 
moment of the force of the spring. 

82 . If the lever of the preceding experiment has not its center 
of gravity on the axis, it will still be necessary to take account of 
the effect produced by the weight of this lever, which cannot, 
practically, be reduced to a simple geometric line. In order to 
determine this we must find the distance of the center of gravity of 
the lever from the axis, and multiply this value by the weight of 
the lever. We then add this product to the moment of the force 
previously obtained. 

Let us suppose, for example, that a weight of 20 grammes, 
suspended at the extremity of a lever arm 200 millimeters long, makes 
equilibrium with the elastic force of a barrel spring. The product 

20 X 200 = 4000 

represents the moment of the force exerted by the weight. 

If, moreover, the weight of the lever is 7 grammes, and the 
distance from its center of gravity to the center of the arbor 143 
millimeters, the moment of the force exerted by the lever will be 

7 X i43 = 1001. 

Adding this value to the moment of the force of the weight, we 
obtain the moment of the force of the spring that we will designate 
by F f then 

F = 4000 -f- 1000 = 5000 grammes 

in round numbers. This is the approximate value of the moment 
of the force of the spring in a watch of 43 millimeters (19 lines). 

Let us remark that generally these levers are furnished with 
counter weights combined in such a manner that the center of 
gravity is found on the axis. 

83 . The number 5000 that we have just obtained, signifies 
that the spring considered is capable of making equilibrium with a 
weight of 5000 grammes suspended at the extremity of a lever arm 
equal to the unit of length, therefore 1 millimeter (Fig. 14). 

84 . Examining in this manner the force of a spring, we will 
prove that it varies very much according to the number of turns 
that it is set up. Experience proves in fact that the moment of the 
force of a spring being, for example, at its maximum point of 
tension, 5000 grammes, this moment constantly diminishes, and 
will not be more than about 3400 grammes when the barrel will 
have executed four rotations around its axis. 


Maintaining or Motive Forces . 


45 


85. We understand then that the imperfections of the primitive 
watches being known, the ancient horologists should have sought 
means for correcting the inequality of the action of the motive 
spring, and that for this purpose they should have invented the 
ingenious arrangement of the fusee , which will be explained 
later on. 

This corrective is really almost entirely abandoned, and is 
seldom used except in marine chronometers ; in pocket watches it 



Fig. 14 


has become useless in proportion as the improvements in the con¬ 
struction of escapements have come into use, and as the isochronism 
of the oscillations of the balance wheel has been obtained. 

Theoretical Study of the Moment of a Spring’s Force. 

86. Coefficient Of Elasticity. When a body receives an exterior 
effort, the molecules which compose it tend to follow the direction 
of this force ; they approach each other or separate themselves, the 
one from the other. The result is a force equal and opposite, 
which tends to make the displaced molecules recover their former 
positions. 

This property, common to all bodies in different degrees, is 
called their elasticity. 










46 


Lessons in Horology. 


According to the effort exerted, the molecules approach or 
leave each other; the first case is an effect of compression or 
contraction, the second is an effect of tension. 

87. The reaction is always equal to the action ; we can then 
measure the elastic force of bodies by the exterior effort which 
is applied to them. The following experiment will explain this 
assertion : 

88. Let us secure one of the extremities of any vertical rod, 
to the other extremity we suspend a weight (Fig. 15). This 

rod from that moment undergoes a certain elonga¬ 



tion, and we can prove that the molecular effort devel¬ 
oped is equal to the weight producing the elongation. 
The elongation of this rod will depend on the size of 
the force P, on the length of the piece in its natural 
state, on the cross section of this piece being assumed 
the same throughout, and finally on the material of 
which it is composed. 

By experimenting on a rubber band we can see 
that under the action of a force P , the transverse section 
of the body diminishes at the same time that the elonga¬ 
tion is produced. This regular diminution on almost 
the entire length of the band does not take place 
uniformly near the two points of fastening. Therefore, 
it is necessary to take the elements for the experiment 
sufficiently removed from these points in order to 
eliminate a source of error which would influence the 
final result. 

Let us take again, for example, a rod of iron, 
whose transverse section is 1 square mm. ; we have 
measured the distance between two marked points 
sufficiently removed from the points of fastening ; let L 
be this length. We suspend from the lower end of 
the rod a weight P , and we measure anew the length 
between the two marks ; we obtain then an elonga- 
Fig. 15 tion l. Experiments made in this manner have de¬ 
monstrated that, provided the load P does not surpass 
a certain limit, / remains proportional to the load. 

Supposing now that the experiment were physically possible, 
let us determine what should be the load P that could produce 
an elongation equal to the original length L. We call this 









Maintaining or Motive Forces. 


47 


particular value of P the coefficient of elasticity of the body ; we 
will designate it by the letter E. 

The elongation being proportional to the load, we have 


whence 



E 

Z’ 


E = 


PL 
l ‘ 


Thus, in the case of an iron rod, whose original length L 
was 1000 millimeters, if we suspended from it a load P of 1000 
grammes, we will find an elongation of 0.05 mm., which gives 

r. 1000 X 1000 

E = — 0.05” = 20000000 gr¬ 
as the coefficient of the elasticity of iron. 

89. The elongation l is inversely proportional to the transverse 
section of the body ; thus for a section of surface the formula 
above will become 


90. When the coefficient of elasticity is known, it is easy to 
determine the value of the force exerted by the molecules of a 
body subjected to the action of an exterior force by the relation 

p _ Els 

^ ~ ~L * 

The fraction ~ represents the elongation per unit of length ; this 
fraction should remain very small for this formula to be exact. 
The quantities E , s , L are constant ; P and / vary together. 

The same formula expresses the relation which connects a 
force P of compression to the contraction /, which results from the 
action of this force, when the piece compressed does not bend. 
We will give then to P and to l the signs -f- and —, -f- for the 
forces of tension and the elongations, — for the forces of com¬ 
pression and the contractions ; the formula then becomes general. 

91. Variation of the Coefficient of Elasticity. All watchmakers 
know that after having forged a piece of brass, the elastic force of 
the metal is increased. In hammering this body one diminishes its 
volume, but one cannot change its weight; the molecules which com¬ 
pose the piece are forced together, and the specific weight of the 
metal will be increased. This simple fact shows us that the coefficient 
of elasticity of solid bodies should vary with their specific weight. 

When a watch (not compensated), regulated to a certain 
temperature, is exposed to a higher temperature, it loses about 



4 8 


Lesso?is in Horology. 


io seconds in 24 hours for each degree centigrade. The spring is 
expanded by the effect of the increase of temperature, its mole¬ 
cules are separated from each other ; the specific weight of the 
metal has diminished at the same time as its coefficient of elasticity. 
The reverse takes place when the watch is observed at a lower 
temperature than that to which it had been regulated. 

It does not appear that the coefficient of elasticity of steel 
undergoes a great variation by the effect of tempering and that of 
reheating. A piece of steel in fact changes its dimensions very 
little by tempering. It has been proven that by tempering in 
water a piece of steel stretches about 3-oVc °f lts original length, 
but that this elongation is lost when the piece is reheated to the 
blue color, the specific weight of the steel not being modified, the 
coefficient of elasticity retains the same value as that which it 
possessed before tempering. 

92. We give here a table of the coefficients of elasticity of 
some bodies employed in horology. The figures below are taken 
from the “ Almanac of the Bureau of Longitudes.”* 


Values of the Coefficients of Elasticity, E. 


Substance. 


Hammer Hardened. 


Steel . 

English Steel. 

{ Very soft 
Demi-soft 
Hard . . 
Silver. 


{ Ordinary . 
Phosphorous 
Laveissiere 

Copper. 

Berry Iron. 

{68 Copper ! ! ! 
f18 Zinc 

German Silver : < 61 Copper 

( 22 Nickel 

Gold . 

Palladium. 

Platinum. 

Iridized Platinum : I 10 pM*. um \ 

1 90 Platinum J 

Flat Glass. 

Zinc. 



19549 
18809 
20705 
20911 
20599 

7358 

7589 

8250 

9061 

12449 

20972 

9277 

9395 

10788 

8132 

II759 

17044 


8735 


Annealed. 

19561 

17278 


7146 


10519 

20794 


5585 

9789 

15518 

21426 

6722 

9292 


Remark— The above values are expressed in kilogrammes ; we will always reduce them 
to grammes whenever we introduce the coefficient of elasticity in our calculations. The 
coefficient of the extra fine steels employed in horology is generally superior to the value given 
in the above table. Experience has led us to employ 23000 for its mean value (therefore 23000000 
grammes. 


* An almanac published by the astronomers of the Paris Observatory. 




















Maintaining or Motive Forces. 


49 


93. Limit Of Elasticity. If we submit any rod to the action of 
different loads, we note that as long as the load does not exceed a 
certain limit, proportionate to the transverse section of the body, 
the rod resumes its original length, after the removal of the weight. 
By increasing the weight so as to pass this limit, the elongation only 
partially disappears or perhaps does not disappear at all. This 
limit is called the ‘ ‘ limit of perfect elasticity ’ ’ of the body con¬ 



sidered. If we continue to increase the weight, the elongation 
becomes more and more apparent and at length a “rupture” is 
produced. 

The limit of perfect elasticity is very slight for certain metals, 
such as lead, red copper, aluminum, etc. Iron, even, does not 
possess a very great limit ; on the other hand, steel, when it is 
tempered, increases its limit of elasticity by suitable reheating. 
This reheating, known under the name of “spring temper,” cor¬ 
responds to the bright blue color. 

It would be of great use in horology to know the exact value 
of this limit of perfect elasticity of hardened and tempered steel ; 
these experiments have not yet been thoroughly studied and the 
data is consequently lacking. For the present we will confine our¬ 
selves to the results with which the practice of horology furnishes us. 

94. Moment of the Elastic Force of a Spring Subjected to a 
Flexion. Let A 0 B 0 be a spring of circular form, of rectangular 










50 


Lessons in Horology . 


section, of thickness e and height h. Let us imagine that this 
blade of spring be divided in the direction of its length into a 
certain number of fibres, one of which, especially, situated in the 
middle of the body, is called the ‘ ‘ neutral fibre * ’ for the reason 
that it does not change its length when the spring undergoes a 
flexion. When this blade is bent in such a way that the radius of 
the neutral fibre diminishes (Fig. 16), the fibres interior to this 
undergo a shortening, while the exterior fibres are lengthened. 

Let ± v be the distance of any fibre from the neutral fibre. 
+ v if the fibre is on the exterior and — v if it is on the interior of 
the neutral fibre. If r 0 represents the radius of the neutral fibre in 
the unchanged position and 0 the angle that the two radii ending 
at the extremities A 0 and B 0 , form between them, we will have the 
length L 0 of the neutral fibre by the equation 

L 0 = r 0 6 

and the length Z/ 0 of any fibre whatever whose radius is r 0 -j- v 
will be 

Bo = (r 0 + v ) 6 . 

If now, one of the extremities of the blade is fastened and we 
bend the other, making it traverse an angle ± a , the radius of the 
neutral fibre will diminish if a is positive, that is to say, if it adds to 
the angle 0. If, on the contrary, the extremity Z? 0 is bent in the 
opposite direction, the angle 0 becomes smaller and we have in 
this case a negative : the radius of the neutral fibre will increase. 

The length L 0 of this fibre has not changed by the flexion ; 
we will have then in this new position, r being the radius of the 
changed position of the fibre, 

Z 0 = r (0 - a), 

we then have 

r 0 0 = r (0 -f a), 

from whence 



The fibre taken whose length is Z/ 0 has become 

L'o — ( r + v) (0 + a). 
Replacing r by the value above, we will have 

L '° = (r+^s + v ) (e + a)> 

and in working out 

L\ = r 0 0 -f-z/O-f'Z'a 




Maintaining or Motive Forces. 


5i 


the elongation / of the fibre considered will be obtained by taking 
the difference between lengths L' and L\ then 

Lf — L' o = /=r o 0-|-z/0-f- v a — r 0 0 — v 0 
and simplifying = v a 

This elongation is positive ; but it will become negative for v 
negative ; that is to say, for the interior fibres there will be a 
shortening. There will also be a shortening if v is positive and a 
negative. If these two values are negative, their product is posi¬ 
tive and we have an elongation. Let us remark that the elonga¬ 
tion / is independent of the radius r 0 of the neutral fibre and that 
consequently the spring can be of any form. 

95. Let us now determine the moment of the force exerted by 
two opposite fibres, situated at equal distances -f- v and — v from 
the neutral fibre and let us suppose that the flexion of the blade 
may have been effected preserving the center O to the changed 
position of the spring ; that is to say, that the blade may have 
taken the position A B (Fig. 16). 

The exterior fibre, which has been lengthened by the flexion, 
will tend to return to its first length and will act with a force P 
whose value is represented by (90) 

Els 


P = 


L 


in which we can replace l by its equivalent v a (94), which gives us 

Evas 


(2) P = 


L 


The interior fibre tends to lengthen out and will exert this same 
force in the opposite direction, therefore 

Evas 


P = 


L 


The moment M P of the simultaneous effort produced by the 
the action of these two fibres wall be equal to the sum of the pro¬ 
ducts of each of the two forces by their respective lever arms r + v 
and r — v. Therefore 

M P — P (r + v) — P (r — v), 
or 

(3) M P— 2 Pv. 

This value is then' independent of the radius of curvature of 
the spring, that is to say of the distance from the exterior attaching 
point of the blade to the center of the barrel. 





52 


Lessons in Horology. 


Replacing in the equation (3), P , by its value (2) we will have 


(4) 


M P = 2 


E a. s v 2 
~L 


Let us now regard the section j, of the fibres considered. 

The cross section of the spring being imagined rectangular, of 
height h and thickness e we will have, supposing first that the blade 



is divided into a definite number of fibres, 10 for example, the 
section of one of these fibres 

s — 0.1 e X h } 

since the thickness of one fibre will be in this case the tenth part of 
the total thickness of the spring. 

Let us admit, what is not absolutely exact, that each separate 
fibre acts through its middle part, that is to say, that the distance v 
from the middle of the fibre nearest to the neutral fibre be 0.05 e 
(Fig. 17), the distance of the middle of the second 0.15 <?, that of 
the third 0.25 for the fourth 0.35 e and then for the fifth 0.45 e. 

Since in the equation (4) the term v is to the second power we 
should raise each of the five preceding values to the square and 


find the sum of 

them. 

We will then have 


1 st fibre 

V — 

0.05 e 

. • . . . v 2 = 0.05 2 

e 2 — 0.0025 el 

2d “ 

V — 

0.15 e 

v 2 = 0.15 2 

e 2 = 0.0225 e 2 

3d “ 

V = 

0.25 e 

V 2 = 0.25 2 

e 2 = 0.0625 e 2 

4th “ 

V = 

0-35 e 

• • • . v 2 = 0.35 2 

e 2 = 0.1225 e 2 

5 th “ 

V — 

0.45 e 

v 2 = 0.45 2 

e 2 — 0.2025 e<1 


The sum of v 2 — 0.4125 e 2 










Maintaining or Motive Forces. 


53 


Replacing now the values determined of j- and of v 2 in the 
equation (4) we will have 


Sum M P 


E a o. 1 eh. 0.4125 e 2 
L 


This formula represents the moment of the force of all the ten 
fibres considered, therefore of the entire spring, while the formula (4) 
gave the value of the moment of two fibres only, the one interior 
and the other exterior, to the neutral fibre. Designating by F the 
preceding sum, we will have, by performing the operations 
indicated 

( 5 ) F 


0.0825 E e z h a 


L 


We have obtained the coefficient 0.0825 by dividing the blade of 
the spring into 10 fibres ; if we had supposed it divided into a 
very large number of fibres, we would have arrived at a value 
very nearly approaching 0.08333, say, y 1 ^. We would have then 
in this case. 


( 6 ) 


F = 


E e 3 h a 
12 L 


96 . We have arrived at this last form, which is the exact 
one, only by approximation. 

Integral calculus furnishes us a means of effecting the above 
calculation with an absolute exactitude and for an infinite number 
of fibres. (*) 

Let us take up again the equation (4) 


M P 


E a s v 2 

17 


Designating the infinitely small thickness of a fibre by d v, we will have for 
the section s 

s = d v. h ; 

on replacing 


d. M P = 2 a -- v 2 dv, 


or 


We will place 


f d - 


M P = 2 


L 

E a h 

~L~ 


Jv 2 dv. 


/• 


and we will have 

F — 2 


d M P = F 
E h a r 


L 


1 Cv 2 d v 


, + c 


(*) We cive below this second manner of calculating the moment of the force of a s P rin K- 
This calculation, as also all that is written in fine print in this treatise, can be omitted by all 
persons unaccustomed to the infinitesimal calculus. 










54 


Lessons in Horology. 


Replacing v by \ <f, that is to say, taking the integral between the limits 
v = \ e and v = o, we will obtain 


F = 


i 

T? 


E h e 3 

~z~ 


a. 


97 . In the preceding formula, a is the angle which we have 
made the free extremity of the spring describe from the position where 
the elastic effort is null, to the point which we wished to study. 
Thus, when we have turned the free extremity of the spring one 
revolution, the original number of revolutions will be increased by 
i, etc. We can then estimate the angle a by counting the number 
of turns which the spring is wound up at the moment considered, 
not forgetting to deduct from this figure the number of turns which 
the spring makes if placed unconfined on a table. Let n be this 
difference, we will have 

a = 2 ir 71 ; 

we can write the formula (6) 

, . ^ E h <? 3 2 rr n 


98 . On calculating the moment of the force of a barrel spring 
by means of this equation, and on then comparing the result 
obtained with that which the experiment gives (83), we generally 
find a slight difference. This difference proceeds essentially from 
the following causes : 

1st. As we have already stated, the value of the coefficient 
of elasticity of the spring with which we are engaged could be 
perceptibly different from that which we have admitted in the 
calculation. 

2d. It is difficult to measure exactly the thickness of the spring : 
a slight error will give a considerable difference in the result. Thus, 
for aspring of 0.18 mm. an error of y-J-g- of a millimeter will 
influence the result one-sixth. 

3d. The transverse section of the blade is rarely a perfect 
rectangle ; the spring is often concave on the outside and convex on 
the inside. 

4th. The calculation supposes the spring to be perfectly free, 
but complicated effects are produced when it is shut up in the barrel. 

When it is wound around the hub of the arbor there is only a 
certain length of the blade which is freed from the coils pressed 
against the drum of the barrel. The moment of the force should 
therefore be calculated according to the length of the blade freed. 




Maintaining or Alotive Forces. 


55 


5th. When the spring is wound up to a certain point, the coils 
of which it is composed deviate from the circular form and spread 
out to one side ; there is thus produced a decomposition of force, 
one of the components of which is directed towards the center of 
the barrel and is transformed into friction. We can add a similar 
defect which is produced at the center and which on combining 
with the exterior fault can diminish, or, in certain cases, increase 
the moment of the force of the spring. 

6th. Considerable friction is produced between the coils of 
the spring ; the oil which we are obliged to use to reduce friction 
produces a slight effect by its adhesive force. 

99. Example for the Numerical Calculation of the Formula (7). 
The dimensions of the spring for a watch 43 mm. diameter (19 lines) 
being the following : 

Thickness, <?, =0.18 mm. 

Height, /z, — 3.6 mm. 

Length, Z, = 650 mm. 

to calculate the moment of the force of this spring. 

When the elastic effort of this spring is nothing, that is to say, 
when it is placed perfectly free on a table, it makes 5 turns. Coiled 
in the interior of the barrel and pressing against the drum, it makes 
14. The development of this spring being 6 turns in the barrel, a 
half turn is given for safety, and we will have, according to what 
has been said, 

n = 14 + 5-5 — 5 = 14.5 turns, 

when the watch is completely wound up. 

Let us take the coefficient of elasticity, E = 23000000. The 
formula (7) can be written thus : 

E h e 3 it n 

aV ■ _ • 


replacing the letters by their values, we have 

__ 23000000 X 3-6 X 0.18 3 X 3- T 4 l6 X H -5 
“ 6 X 650 

Effecting the above calculations we find that 

F — 5640 gr. 

for n = 14.5 turns. 





56 


Lessons in Horology. 


The simplest way of effecting the above calculation is by means 
of logarithims. We give below the method of such an operation : 

Log. E = log. 23000000 = 7.3617278 

+ log. <? 3 = log. 0.18 3 = 0.7658175 — 3 

+ log. h = log. 3.6 = 0.5563025 

+ log. it = log. 3.1416 = O.4971499 

Log. numerator 6.1809977. 

Log. L = log. 650 = 2.8129134 
+ log. 6 = 0.7781513 

Log. denominator 3.5910647. 

Log. numerator 6.1809977 
— log. denominator 3.5910647 

2.5899330 = log. 388.985. 

-f log. n = log. 14.5 — 1.1613680 

3.7513010 = log. 5640. 

The preceding calculation shows that the moment of the force 
of the spring is 388.985 gr. for a number n — 1. For n — 14.5 
it is 5640 gr. When, on account of the running of the watch, the 
barrel has made one turn, n will have diminished one unit and will 
only have 13.5 as value ; the moment of the force of the spring 
will have diminished 388.985 gr., or, in round numbers, 389 gr. 
We can then form the following table : 

F for n = 14.5 = 5640, the spring is entirely wound up. 

Fx “ n = 13.5 = 5251, the barrel has made one turn. 

F x “ n = 12.5 = 4862, the barrel has made two turns. 

Fz “ n = 11.5 = 4473, the barrel has made three turns. 

F i “ n — 10.5 — 4084, the barrel has made four turns. 

100. Inequality of the Elastic Force of the Spring. The 

moment F of the force of a spring is then greater when the watch 
is completely wound than when it is about to stop. This fact has 
been already demonstrated to us by experiments (84). 

It is necessary to confine this inequality of the motive force 
within the narrowest limits possible. Let us note for this purpose 
that in the numerical calculation of the formula (7) we have succes¬ 
sively replaced n by n — 1, n — 2, n — 3, and n — 4 ; in this last 
case the watch is at the instant when it is about to stop, if the barrel 
is furnished with stop works. But the ratio between n and n — 4 
being greater as n is smaller, it will be proper, in order to diminish 
the inequality of the force, to give to the number of turns, n , of 






Maintaining or Motive Forces. 


57 


the spring, as great a value as possible. The following demonstra¬ 
tion will better explain this idea. 

101. Since we can use springs producing the same initial moment 
of force F 0 , but whose dimensions and number of turns, ?i, are 
different, we understand that F 4 may vary in certain cases much 
less than in others. 

Let us suppose, for example, two springs producing, when 
wound up, the same moment of force F 0 — 4000 ; the first having a 
number of turns n = 10, the second a number n — 20. For the 
first we would have in the formula (7) the value 


, E e z h 2 tr 

TJ 7 — 4°°. 


and for the second this same value will be 


j E e z h 2 it 
TY ~ L ~ 


200 . 


When the two barrels will have executed one revolution, the num¬ 
ber of turns of the first will be n 1 = 9 and that of second n Y = 19. 
We will then have successively : 


First Case. 


F 0 = 400 X 
F 1 — 400 X 
F 2 = 400 X 
F z — 400 X 
F± = 400 X 


10 = 4000 
9 == 3600 
8 = 3200 
7 = 2800 
6 = 2400 


Second Case. 

F 0 = 200 X 20 = 4000 
F l = 200 X 19 = 3800 
F 2 — 200 X 18 = 3600 
F s = 200 X 17 — 3400 
F± -- 200 X 16 — 3200 


It is thus easy to see that the moment of the force diminishes 
in the first case much more rapidly than in the second, and that the 
force of the second spring approaches much more a constant value 
than that of the first. 

It is best, then, to give to the number of turns, n , the greatest 
value possible. 

For a given spring this number cannot, however, exceed a cer¬ 
tain value determined by the limit of perfect elasticity of the steel, 
which cannot be exceeded without setting or breaking the spring. 

This limit depends on the elongation per unit of length ~ of the 
exterior fibre. We have had (94) 

l — v a 

which can be written 

/ _ v a 

~L ~ ~~L T 







5 « 


Lessons in Horology. 


In the preceding numerical example we have had the following 
values : 

v — \ e — 0.09, a = 2 ir n — 14.5 X 2 ir, L = 650 mm. 

We will then have 

l _ 0.09 X i 4-5 X 2 tt 
L ~ 650 

and on performing the calculations 

—j- — 0.012614 mm. 

We can admit this value of as the limit allowing sufficient 
security, and established by practice. 

102 . It must not be forgotten, however, that the nature of the 
steel, the manner in which the springs are manufactured, hardened 
and tempered, can materially modify this limit. 

The springs are hardened and tempered in circular form, with 
about 100 mm. radius ; they are then worked in a spring tool and 
by this operation the fibres undergo quite an unequal elongation, 
since, from the first circular form, they pass to a spiral form whose 
radii of curvature for the interior coils are much smaller than thoss 
for the exterior coils. 

It follows, from this operation, that the exterior fibres are 
elongated more in the inner coils than in the outer ones. It is 
for this reason that the springs break more often interiorly than 
exteriorly. 

103 . The form which a spring has on leaving the hands of the 
maker is very variable and it can be with difficulty represented by 
a general equation. 

This primitive form is not preserved after the watch has run a 
certain time : the spring 4 4 gives ’ ’ a little at first and finally ends by 
taking a form which it keeps permanently. This last form is the one 
which should be taken as the starting point from which to determine 
the angle 2 ir n giving the degree of flexion in the formula (7). 
Starting from this position, we can admit that the elongation per 
unit of weight -j- is much the same for the whole length of the spring. 

104 . Length Of the Spring. Since we cannot unfold a spring, 
in order to measure its length, without modifying its interior struc¬ 
ture, it is convenient to have at our disposal a simple formula 
enabling us to calculate the value L. 

Supposing the spring coiled in the interior of the barrel ; we 
will admit that the radius extending to the interior coils may be 




Maintaining or Motive Forces. 


59 


equal to f R in the position of the spring at rest, R being the 
interior radius of the barrel. We can, without great error, substi¬ 
tute circumferences for coils and obtain the length of the spring by 
multiplying the mean radius, 

2 — S A > 

by 2 v N\ on deciding to designate by N' the number of coils in the 
spring when it is pressed against the side of the barrel. To this value 
must be added the length of the end of the spring which is detached 
from the coils so as to be hooked to the hub, which is about 

2 it x — = ir R ; 

2 

we would have, therefore, the length 

L ■= 2 ir R ( | N' + 1 ). 

We have, for example, in the calculation of the length of a 
spring for a watch of 43 mm. the following values : 

R — 8.8 mm. and N / = 13 coils. 

Replacing the values, we will have 


whence 


L — 2 X 3.1416 X 8.8 (f x 13 + 1); 

L = 626 mm. 


Development of a Spring. 

105. When a watch spring is put in the barrel it is wound on 
itself and forms a certain number of coils, the outside one of which 
presses against the side of the barrel and the succeeding ones against 
each other, taking the form of a spiral of Archimedes, with grada¬ 
tion equal to the thickness of the spring. The inner end of the 
blade is disengaged abruptly from the coils and is fastened to the 
hub of the arbor. 

In order to “wind” the spring, we can hold the barrel and 
turn the arbor several turns until the spring may be entirely wound 
around the hub, with the exception of its outer end, which remains 
fastened to the side of the barrel. 

It is evident that the number Wof revolutions which the arbor 
has been able to make is equal to the difference between the num¬ 
ber of coils that the spring has in these two extreme positions. 




6o 


Lessons in Horology. 


Let N\ be the number of coils which the spring makes when it 
is pressed against the side of the barrel ; N" the number of coils 
when wound around the hub. We will have then JV= N" — N'. 

In order to simplify the calculations, let us neglect the inner 
and outer ends of the blade which are disengaged from the coils, and 



Fig. 18 Fig 19. 


admit that the space occupied by the spring in the two positions is 
a cylindrical volume. Let us designate, moreover by 
R , the interior radius of the barrel ; 
r, the radius of the hub ; 

r ', the radius extending to the interior coil of the spring when 
it is pressed against the side of the barrel (Fig. 18) ; 
r", the radius extending to the outer coil of the spring when it 
is wound around the hub (Fig. 19) ; 

<?, the thickness of the spring. 

We can w r rite 


N" = —-- and N’ = — 


and consequently 


(1) N — N" — N' = 


.// 


r' 

—> 

R — r' 

e 


We can find the value of r” in functions of R, r* and r on 
observing that the surfaces occupied by the spring in the two posi¬ 
tions are equivalent. 

When it is pressed against the side of the barrel, this surface is 


S = * (R 2 — r' 2 ), 
















Maintaining or Motive Forces. 


61 


and when it is wound around the hub 


Therefore, 
from whence 


6 * = <ir ( r " 2 — r 2 ). 
r " 2 — r 2 = R 2 — r /2 , 
r" = ;/ R 2 — r ' 2 -f r 2 . 


Substituting this value of r” in the equation (1), we have 


(2) 


N = 



R 2 — ? V2 4 - r 2 — r — R -\- 



106. Let us note that when we obtain 


equation 


N' = 


R 


the value of e in the 


e 

we find that it differs from the real value, which is always less. This 
difference arises from the fact that on account of certain inequalities 
of the spring, the coils of the blade do not strictly superpose. 

10T. Maximum of N in Terms of r'. The equation (2) 
indicates that for a barrel whose interior radius R, the radius of the 
hub r and the thickness of the spring e, are determined, the num¬ 
ber of turns of development N varies with the radius ?' extending 
to the first inner coil of the spring in its state of rest. This last 
radius, r', depends on the length of the blade. 

Let us apply, in the first place, this formula to a numerical 
example and use the following data : 

Interior radius of the barrel R = 3 

Radius of the hub. r = 1 

Fraction -. = 13. 

e ^ 

The equation (2) will become, after replacing the known 
quantities by their values, 



The smallest value that the radius r* can have is 

r' = r — 1 


and its greatest value may be 

r' = R = 3. 

In the first case the spring has a number of coils sufficient to com¬ 
pletely fill the space between the side of the barrel and the hub, and 
in the second this number of coils is nothing. The reasoning shows 









62 


Lessons in Horology. 


that in these two extreme cases the development of the spring would 
be nothing, which the application of the formula (2) also proves. 

Replacing successively in this same formula r' by 1.1, 1.2, 
1.3, etc., up to 2.9, we can form the following table : 


r 1 

N 

r f 

N 

r' 

N 

1 

0 

i -7 

4.76 

2.4 

596 

1.1 

0.84 

1.8 

5-2 

2.5 

567 

1.2 

1.63 

i -9 

5-56 

2.6 

5-2 

i -3 

2-37 

2 

5 -84 

2.7 

4-5 

1.4 

3.06 

2.1 

6.03 

2.8 

3-5 

1-5 

3-69 

2.2 

6.13 

2.9 

2.09 

1.6 

4.26 

2.3 

6.11 

3 

0 


One sees that the maximum number of coils in the develop¬ 
ment of the spring is given by a radius r' equal to about 2.2. 

In practice, f of R has been adopted as the value of r' for the 
reason that the regularity of the power from beginning to end 
increases with the length of the spring. 

108 . The calculus enables one to determine the exact value of the 
maximum of N in function of r '. 

Let us take up again the equation (2) : 

H ~ (1 /R 2 — r' 2 -| - r 2 — r — R -j- r / ), 

in which the two variables are N and r'. Let us differentiate this equation 
by placing 

R2 — r /2 q_ r 2 — 

we will have 

d z — — 2 r' d r' 

and 

d H — j (l z* d z + d r' )> 

Replacing z and d z by their values, it becomes 































Maintaining or Motive Forces. 


63 


equating this derivative to zero : 


which gives 


and 


K- 


r• 


\/R 2 — r ' 2 -f- 




+ 1 


)=o. 


V R 2 — r ' 2 + r 2 


= 


r' — ± \/r 2 — r ' 2 + r 2 
Raising to the square we have 


• /2 


and 


R 2 — r ' 2 -J- r2 


(3) r f 


-V 


j ? 2 + r 2 


Substituting this value of r' in the equation 


.// 


we will have 


V R 2 — r ' 2 + 


.// 


- 


R 2 + r 2 


yl 


= 4 


R 2 + r 2 


We see that the maximum of N occurs when one has 


(4) r' = r" = 

Since it is the custom, in practice, to make the radius of the hub equal 
to one-third of the interior radius of the barrel, we can place 

R = 3 and r = 1 

and we will have 

r ' = r " = y / - 9 = V~s = 2.236. 

109 . In order to represent graphically, the equation (2), let us 
refer it to two rectangular coordinate axes (Fig. 20) and lay off on 
the axis of the abscissa the values of r and on the axis of the 
ordinate the corresponding values of IV, and connect the points 
thus obtained by a curved line. 

One sees that on making the unoccupied space of the barrel 
equal to the part occupied by the spring one does not obtain the 
maximum turns of development of the spring. 

110. If one divides the interior of the barrel, giving 
^ R to the part occupied by the spring, 

£ R for the empty space, 

\ R for the radius of the hub, 
one has 


7 (1/3 2 — 2 2 + I — i — 3 + 2) 



















6 4 


Lessons in Horology. 


and 

N = 7 W r ~ 2 ) = 7< 2 '44 - *) = i X 0.44.* 
Consequently, if 

~ = 13 one will have N — 5.7 turns 

i = 14 “ “ “ N= 6.1 “ 

7 = 15 “ “ “ N= 6.6 “ 

e ^ 

111. Diameter Of the Hub. The custom of making the radius 


81 . 



of the hub equal to one-third of the interior radius of the barrel has 
been established by long practice. When one leaves the hub 
greater than this value, one does not obtain a sufficient number of 
turns of development of the spring, and when one makes it smaller 
the spring is apt to break, or if it is too much reduced in temper it 

*It is best not to extend the figures of the square root of 6, because on account of the 
interior part of the spring being attached to the hub, one therefore loses more readily a little 
of the development. 




































Maintaining or Motive Forces . 


65 


may ‘ ‘ set. ’ ’ Cases present themselves, however, where, in a given 
barrel, one desires to introduce a thinner spring than that which 
one has been in the habit of using and it may then be asked if 
one could not reduce the diameter of the hub, in order to obtain 
a greater development of the spring. Let us then examine this 
question. 

It is known that when one submits a rod of steel to a tension, 
the transverse section being equal to one square millimeter, this 
rod lengthens, and that when the load which produces this elonga¬ 
tion reaches a certain value, a rupture is produced. Now, expe¬ 
riments made with spring wire have demonstrated that the wire 
breaks if the weight reaches a value of 135 kilogrammes. 

Let us note that the wire which was used for this experiment 
was not hardened. 

Let us first seek again the elongation / per unit of length which 
the rod underwent at the instant immediately preceding the rupture. 

We have (88) 


Taking £ = 23000000, P = 135000 gr., L— 1, we will place 


whence 


_ 135000 _ 135 

23000000 23000 

/ = 0.0058695 mm. 


On the other hand, let us calculate the elongation of the exterior 
fibre of a barrel spring as it is admitted in practice and compare the 
two results. For this purpose let us take the spring of a watch of 
43 mm. (19 lines) which has furnished proof of being able to bear 
the desired flexion. 

We will have the elongation per unit of length from the 
formula (94) 


The thickness of this spring is 0.18 mm., consequently 


v =■ 0.09 mm. 

In the interior of the barrel and pressed against the drum this 
spring had 13 turns and 5 outside of the barrel ; moreover, after 
being wound, it was set up 5.5 turns. With this information we find 


a = (13 — 5 + 5 5) X 2 ir = 27 it. 






66 


Lessons in Horology . 


The length of the spring is 600 mm.; one will have consequently 

, 0.09 X 27 X 3.1416 

l — — ? 0 -**• — = 0.01272. 

600 

Comparing the above figures, 

elongation by tension, l — 0.0058695, 
elongation by flexion, l = 0.01272, 

we can establish the astonishing result that the exterior fibres of a 
spring can sustain an elongation per unit of length twice as great as 
that which produces a rupture by tension. 

This fact cannot be explained by the supposition of a superior 
quality of steel to that of the metal composing the rod which broke 
under the action of a weight of 135 kilog.; because this last steel 
was certainly of the first quality. It must then be admitted that 
the exterior fibre of a spring does not break, when the elongation 
that it acquires is equal to that which produces rupture by tension, 
for the reason that it is retained by the interior fibres. 

In the presence of this fact one can admit as the limit of 
elongation that the exterior fibre of a spring can bear without 

breaking, the value . 

0 l = 0.012 mm. 

One remains within this limit in making the diameter of the hub 
equal to one-third of the interior diameter of the barrel and in using 
a spring making 13 turns in one-third of the interior radius of this 
same barrel. 

When one desires to use a thicker or thinner spring, one must 
in this case determine the diameter of the hub with relation to the 
thickness chosen. Thus the interior diameter of the barrel which 
we have used in the preceding experiment being 17.4 mm., the 
diameter of the hub was then 

^ 7*4 _ Q 

- — 5.0 mm. 

3 

Dividing this diameter by the thickness of the spring 0.18 one 
arrives at the conclusion that the diameter of the hub should be 32 
times the thickness of the spring, in round numbers. If the 
diameter of the hub is smaller than this proportion, the spring runs 
the risk of breaking or, if it is too soft, of setting. 

Work Produced by a Spring. 

112 . One determines the mechanical work which a spring pro¬ 
duces at each oscillation of the balance wheel, by dividing the work 




Maintaining or Motive Forces. 


67 


displayed by the spring during one turn of the barrel, by the num¬ 
ber of oscillations made by the balance wheel during this time. 

Let F — 4800 be the moment of the force of the spring of a 
watch whose balance executes 18000 oscillations an hour ; we will 
obtain the mechanical work effected by the spring during one turn 
of the barrel by the product 

W — 4800 X 2 rr. 

If the barrel of 80 teeth gears into the center pinion of 10 leaves, 
the number of corresponding oscillations will be 

^°- X 18000; 

10 

the mechanical work during one oscillation will then be 


Wm = 


4800 X 2 ^ 
f5 X 18000 


0.2094 gr.mm. 


If, on the other hand, the watch only beats 16200 oscillations 
per hour, one would have in this case 


Wm 


4800 X 2 ir 
~ X 16200 


0.2327 gr.mm. 


One sees then that the work of the force received by the balance 
wheel at each oscillation is increased by diminishing the number of 
these oscillations. Let us suppose again that the balance wheel 
executes 18000 oscillations, but that the pinion of the center wheel 
has 12 leaves in place of 10, one would then have 


4800 X 2 ir 

Wm = so . . q = 0.251328 gr.mm. 

12 X 18000 

One, consequently, also increases the force by diminishing the 
duration of a revolution of the barrel. 


The Fusee 

113 . We know that the law of the variations in the action of a 
spring which unwinds, is complex, and that the force developed has 
widely separated limits, for a spring of the same thickness its 
whole length. 

In order to remedy this defect, the ingenious arrangement of 
the fusee was conceived long ago, consisting of a solid body whose 





68 


Lessons in Horology. 


sectional revolution is very nearly parabolic, and whose surface is 
grooved with a helicoidal curve. This piece is mounted on the 
axis of a toothed wheel A (Fig. 21) gearing with the pinion of the 
center wheel. The teeth of the barrel are, in this case, suppressed, 
and its arbor remains constantly fixed. 


Fig. 21 

This arrangement permits of the complete equalization of the 
motive action of the spring. In effect, when one has just wound 
the watch, the spring is completely coiled ; a steel chain, one end 
of which is hooked to the fusee and the other to the barrel, is at this 
moment almost completely wrapped around the spiral lines of the 
axis of the fusee. On unwinding itself, the spring turns the barrel, 
which communicates its movement to the fusee by the intermediary 
of the chain. This unwraps itself, little by little, from the fusee, 
and wraps around the barrel until there remain no more turns on the 
fusee. It is evident that if the tension of the spring continually 
diminishes, this action works in the contrary sense, always at a 
greater distance from the axis of the fusee. The motive work, 
product of the tension by the distance traversed, gives, designating 
the tension by P, the distance to the axis by r and the speed of 
rotation by w , 

P r w. 

This work should be constant if the speed of the wheel fixed 
on the axis of the fusee is constant, that is to say, if there is a 
uniform angular speed, and the fusee is grooved in such a manner 
that the product P r remains constant. The variations of P must 
then be the reverse of those of r. An exception, however, must be 
made to the preceding, if one takes into account the friction of the 
pivots in the plates ; this friction, in fact, diminishes constantly as 












































































Maintaining or Motive Forces. 


69 


the pressure diminishes. We will, however, neglect this factor, in 
order not to complicate the following theory. 

114. To determine practically whether or not the variation of 
the force of a spring is exactly counterbalanced by the form of the 
fusee, one uses a lever and a weight, as we have seen before (81) ; 
one fixes the lever on the square of the arbor of the fusee ; the form 
of this piece will be exact if the weight carried by the lever makes 
equilibrium with the force of the spring at the same distance from 
the axis for each point of the successive rotations of the fusee. 

115. Calculation of the Variable Radius of the Fusee's Helix. Let 
R , be the interior radius of the barrel, half the thickness of the 

chain being included therein ; 

r, the variable radius of the fusee; 
r 0 , its initial radius (in rand r 0 is also included half the thickness 
of the chain) ; 

G, the maximum angle which the spring is wound, starting from the 
position where the elastic effect is null, and corresponding 
to the instant when the chain acts at the extremity of the 
radius r 0 ; 

a, the angle which the barrel has turned, starting from the posi¬ 
tion 0 ; 

p, the angle which the fusee has turned, starting from the instant 
when the chain acts at the extremity of the radius r 0 . 

The moment of the force of the spring can be expressed by 



tV F F h 

(0 - 

Placing 

F - ‘ L 

T \ Ee 3 h 


= M 


L 

we will have 

F = M (0 

- ^). 


The force F' acting at the exterior of the barrel should be 


F' = 


M 

R 



) 


and the moment F 1 with relation to the axis of the fusee is 


The values of r and of a should vary in such a manner that, in 
order that F x may be constant, we should have a equal to zero for 
r equal to r 0 ; we will then have 






70 


Lessons in Horology. 


M r n Mr 


whence 


R 


R 


(0-a), 


<*> r = ftri 


When the chain wraps up an infinitely small quantity, R d o, 
on the barrel, it unwraps the same length, r d p, from the fusee. 


One has then 


R d a = r d p, 


but, because of the equation (i), 

d P = ( 0 — a ) da. 

r ° w 

On integrating, it becomes 

p = 7Te/ (e-aMa = 77e 0 9rfa "/ arfa )' 

These integrals should be taken between the limits a = o and 
a = a ', one will have 


consequently, 


j^0 d a = 0 a and j^a d a = \ a 2 ; 

P = rf t ( 9 ° ~ * °*>- 


Drawing from this equation the value of a one will have first 


1 


• • = ^ P; 


changing the signs, adding 0 2 to each member and multiplying by 2, 
it becomes 


2 e a + 0 2 = — — % - e p + 0 2 , 


and 


consequently, 


R 




P + «•; 


0 = 0 * e \l 1 - irk p - 

Replacing now in equation (1) a by this value, we will obtain 

^0 6 


r = 


-e ±A Ji 


2 r o 

R 0 


P 


r — 


V 


I - - 2 -^- p 

R 0 p - 


or 




















Maintaining or Motive Forces. 


7i 


Placing 0 = 2 ix n and p — 2 it n' we will have, finally, 


(2) 



116. Numerical Calculation of the Preceding Equation. Let 

R = 8 mm., r 0 = 5 mm., 0 = 12 X 2ir and let us calculate first 
the value of the radius r for an angle p = 2 ir; we will write, on 
replacing values, 


r 1 = — 

V 

the calculation gives : 


1 


2X5 

8X12 



10 * 
96 


Log : 86 = 1.9344985 
— log : 96 = 1.9822712 


o. 6989700 


Log : J p = 0.9761136 — 


1.9522273 — 2 

I 


Log :_5_ = 

— log : - J||= 0.9761136 

Log : r 1 = 0.7228564 
and r 1 = 5.2827^. 


Successive calculations will give us in an analogous manner the 
following results, which we group in a table : 


For 

P = 

2 

IT) 

r f 

5-2827. 

4 < 

P = 

4 

IT, 

r" = 

5 - 6195 - 

< < 

P = 

6 


yltl _ 

6.0302. 

i < 

P = 

8 

TO 

r"" _ 

6.5491. 

4 4 

P = 

10 

TT, 

yltltt _ 

7.2231, etc. 


117. Other Calculations. It more often happens, in practice, 
that one is given the greatest radius of the fusee, and that the ques¬ 
tion is to determine the variable radius, starting from this value. 
This problem, the inverse of the preceding, is solved in an 
analogous manner. Preserving the same notations as in the pre¬ 
ceding case, let r 0 in this case be the greatest radius of the fusee 
and 0 the angle which the spring is set up at the instant when the 
chain acts at the extremity of the radius r 0 of the fusee. We have 
the moment of the force of the spring : 

F 0 = M 0, 

and in die initial position, when the barrel has turned an angle a, 

F — M (Q -f a). 












72 


Lessons in Horology . 


The force F' acting at the exterior of the barrel will be for the 
two cases : 

F\ = ~ 6 and F' = ~ (6 + a). 

and the moment of these forces with relation to the axis of the 
fusee will be : 

F" 0 = e and F" = ~ ( 6 + a). 

Making these two values equal, one has 

Mr ° 0 = M J- (0 + a), 


R 


R 


or 


r 0 0 = r (0 + a), 


from whence one extracts 

<-> ' = e'Va- 

As in the preceding case, we place 

r d p — R d a, 

from whence 

d J3 =: . (j a, 

v 

Replacing r by its value (i), one obtains 

d P = ~-Q ( 0 + a) d a, 

and on integrating, 

p = ?re‘ (/ erfa+ J arfa > 


from whence 


R 


P — g- (0a + | a 2 ). 


Transformations analogous to the preceding case will give us 
successively : 

!•’ + •* = 'y 9 - p, 


a 2 -f- 2 0 a -j- 0* 


r 0 e 
R 


P + 0 2 , 


- V 


a + 0 = 

0 " V 


r 0 e 

R 


P+ 6 2 , 


R 


P + 0 2 , 


a =3 


- e ± e ^ 

















Maintaining or Motive Forces. 


73 


The value of a, extracted from the equation (i), is equal to 


consequently, 


and 


from whence 


a = To e _ e, 

r 


— e — e = — e ± e FFo o , 
r \ r e p + I# 


r 


5 - 0 


A I 2 r 0 P i T 

6 + X ' 


r — 


J-L2l p + 

\ I\ 0 ^ ’ 


or still further, by substituting p = 2 -n- n f and o = 2 * n, 

(2) r = — - 

IT + 


118. Numerical Calculation of the Preceding Equation. As an 

example of the application of the preceding calculation, let us 
determine the dimensions of a marine chronometer’s fusee and let 
the following be the data : 

Exterior radius of the barrel including half the thickness of 

the chain,. R = 21.7 mm. 

Maximum radius of the fusee, . . . . r 0 = 18.3 mm. 

Development of the spring,. n — 3.4 turns. 

Let us admit, that when the spring is set up one turn, the chain 
acts on an angle p = o, in this case then 

n' = o. 


When the fusee has made one turn, we then will have n' = 1, 
and on replacing the letters by their values in the formula (2), we 
will have 




18.3 


2 X 18.3 X 1 
\ 21.7 X 34 


+ 1. 


The calculation gives 


a T 

Log : (2 X 18.3) = log : 36.6 = 1.5634811 Log : 21.7 = 1.3364597. 

— log : (27.7X 34) = 1.8679386 + log : 3.4 = 0.5314789. 


0.6955425 — 1 

Corresponding number = 0.4961 


1.8679386. 





















74 


Lessons in Horology. 


x^og : 



0.4961 -j- 1 


log: 1.4961 
2 


Log: 1.4961 = 0.1749606 Log: 18.3 — 1.2624511 
log: j/ 1.4961 = 0.0874803 — 0.0874803 

,, 1.1749708 

Number = 14.961 


We will then have the radius of the fusee for a number of 


turns n' — 1 : 


r x = 14.961 mm. 


Replacing, successively, in the preceding formula ri by 2, 3, 4, 
etc., one will arrive at the following results : 


For 

n' -= 0, 

r o = 

I8.3 

4 4 

ri — 1, 

*T = 

I4.961 

4 4 

n! — 2, 

r 2 = 

I2.965 

4 < 

ri = 3, 

r z = 

I I.601 

4 < 

ri = 4, 

r 4 = 

10.593 

4 4 

n * = 5, 

^5 = 

9.809 

4 4 

n! = 6, 

^6 = 

9.177 

4 4 

ri — 7» 

^7 = 

8.653 

4 4 

ri = 8, 

= 

8. 21 


119. Uniformity of the Spring's Force in Fusee Watches. In 

order to obtain perfect uniformity of the spring’s force in fusee 
watches, it is not sufficient alone to construct the fusee in a manner 
conformed to the data of the preceding calculations. There are 
other factors which must be taken into account, and about which 
we will give some explanations. 

In order to verify the uniformity of the force of the spring with 
relation to the fusee, or, to speak in shop parlance, in order to 
equalize the fusee, one places the fusee and the barrel between the 
two plates of the watch, puts the chain in place, sets up the spring, 
and fastens on the arbor of the fusee the lever that we have 
mentioned (114). 

Holding the movement of the watch in the hand, one then 
turns the lever a quarter of a turn and establishes equilibrium by 
means of weights ; then one turns the lever 1, 2, 3, etc., turns, 
taking care to notice if at each revolution the equilibrium is 
maintained. 

But practice teaches that if, in this operation, one finds an 
increase of force, one approaches perfect equality by further setting 
up the mainspring ; if, on the contrary, the instrument shows a 









Maintaining or Motive Forces. 


75 


decrease of force, one lets the spring dow?i. When one nas found 
the uniformity of the force, the initial position of the spring’s 
tension is preserved by marking a point on the pivot of the 
barrel arbor, and by repeating this point on the plate of the watch 
opposite to the position that the former occupies. 

Let us remark that it is necessary to renew this operation each 
time that one replaces the spring experimented with, by a new spring. 

120. Let us seek now for an explanation of the effect which is 
produced in the preceding experiments. Let us admit that the 
radii of the spiral lines of the fusee may have been calculated for a 
spring which, being wound, is set up io turns and has unwound 
3 turns at the moment when the chain is completely unrolled from 
the fusee. 

Being completely wound, the moment of the force of the 

spring is then proportional to.io. 

When the barrel has made i turn, this moment is proportional to 9. 
When the barrel has made 2 turns, this moment is proportional to 8. 
When the barrel has made 3 turns, this moment is proportional to 7. 
From the top to the bottom, the moment of the force has 

diminished =. 0.3, 

and in order that equilibrium may be produced, the radii of the 
spiral lines of the fusee must have increased in the same proportion. 

If, on the other hand, the spring was set up only 9 turns, the 
chain being wound on the fusee, we would have in this case : 

Spring completely stretched, moment of the force proportional to 9. 
The barrel has made 1 turn, moment of the force proportional to 8. 
The barrel has made 2 turns, moment of the force proportional to 7. 
The barrel has made 3 turns, moment of the force proportional to 6. 

The moment of the force has then diminished f — £ = 0.333. 
This decrease is superior to that of the first case, the same fusee will 
not produce equilibrium and one sees thus that it will be necessary, 
in order to have equilibrium, to set up the spring another turn. 

121 . We possess in this way a means of regulating, practically, 
the moment of the force of the spring with relation to the axis of 
the fusee, taking into account certain factors which have not been 
introduced into the preceding calculations ; the principal among 
them being friction. 

It is evident that on setting the spring up further, we increase 
its energy ; it is, moreover, only in rare cases that any incon¬ 
venience will result from this increase of force. 




76 


Lessons in Horology , 


122. In establishing the theory of the barrel spring, we have 
admitted these springs to be of the same thickness from one end to 
the other, that their coiled blades remain always concentric, that is 
to say, retain a spiral form during their development ; and, finally, 
that they are always free. 

Practice shows that these conditions are not fulfilled by the 
spring inclosed in a barrel and that they cannot be so except for 
a free spring, such as the hairspring. 

Let us examine rapidly, how r ever, these three facts, commenc¬ 
ing with the last expressed. 

123. Complete liberty of the blades of the spring does not 
exist practically. Let us suppose, in effect, that one has turned 
the barrel arbor a small angle, a quarter turn, for example. In this 
position all the coils of the spring have not yet entered into play : 
there are found a certain number which remain pressed against each 
other, forming part of the barrel. One should then take the acting 
length of the spring equal to the length of the spring which has 
become free, and the number n equal to the number of turns that 
this free part contains at the instant considered, diminishing this 
value by the number of turns that this same length possesses when 

it is placed freely on the 
table (97). 

Thus (Fig. 22), if a 
is the end of the spring 
hooked to the hub, and 
b the point at which the 
free coils become sepa¬ 
rated from those which 
remain pressed against 
the drum of the barrel, 
the length L in the 
equation (97 ) 

„ E h <? 3 2 it n 
F = iTz 

is then only equal to the 
length of the part a b of 
the spring. Furthermore, the number n will be equal to the num¬ 
ber ot coils which this part of the blade contains, diminished by the 
number of coils which this same length contains in the free posi¬ 
tion of the spring (Fig. 23). For example, we have (Fig. 22) 



Fig. 22 











Maintaining or Motive Forces. 


77 


a b — 2f coils and (Fig. 23) a b — coils ; consequently, 
n — 2f - 2\ = i. 

Let us note that in most watches in which the barrel makes 
three turns in 24 hours, one can readily admit that the whole blade 



Fig. 23 


becomes free, while the spring is developed these three turns, and 
one can employ, without great error, the above formula, without 
modifying anything therein. 

124. Some have tried to use springs of varying thickness, that 
is, those whose blades increase or diminish in thickness from one end 
to the other. 

Let us suppose in the first place that the thickness may con¬ 
stantly increase ; at the interior, the blade, being thinner, will bend 
more easily ; during the winding of the watch, the coils which 
detach themselves from those which remain pressed together, will 
perform this movement in a more gradual manner than if the 




78 


Lessons in Horology. 


thickness were the same along the whole length. On continuing to 
wind the spring some of the coils will be wrapped around the hub 
and form part of it. Thus, the moment of the force of the spring 
could only be determined in this case by taking the length of its 
free part alone, and the value of n should also be determined accord¬ 
ing to this length. Since the thickness, moreover, is variable, it 
becomes difficult to determine by calculation the force of such a 
spring in a sufficiently exact manner. 

Such springs are used to advantage in fusee watches, because 
they have a more concentric development and consequently produce 
less friction between the coils. 

If, on account of the diminution of thickness at the interior, 
there results a greater difference between the moments of force of 
the beginning and the end, this difference could easily be corrected 
by the fusee. 

125. Springs thicker at the' interior than at the exterior are 
hardly to be recommended, for the interior part of the blade bending 
only with difficulty is hard to wind around the hub ; it has, moreover, 
the effect of detaching a greater length of blade from the part that 
remains pressed against the barrel, which produces considerable 
friction between the coils. These springs have, moreover, a great 
tendency to break. 

126. The principal defect of the development of the spring in 
the interior of the barrel is that which arises from the eccentric 
coiling or uncoiling of the blades ; these push themselves to one 
side, both at the interior and at the exterior of the spring. This is, 
also, an analogous fact to that which shows itself in a flat spiral 
without return curve. When the interior fault comes into contact 
with the exterior fault, the spring makes a sudden jump, producing 
a noise well known to watchmakers. The exterior fault can be 
remedied by fixing to the spring a flexible check of sufficient length, 
about a half turn ; this check should be made thin in the part 
which is fastened to the barrel, in order to permit it to follow freely 
the coiling up of the spring. A better remedy would be to have 
the last exterior half turn thicker and make it gradually thinner to 
suit the conditions. 

Stop-Work. 

127. We designate by this name a mechanism fastened to the 
barrel and whose object is to stop the winding before the spring is 
completely coiled around the hub. This same mechanism also 


Maintaining or Motive Forces. 


79 



stops the running of the watch before the spring is completely 
pressed against the inner wall of the barrel ; its effect then is to 
utilize only a part of the development of the spring, that during 
which the force is most equal. Thus the total development of the 
spring being, for example, six turns, if the unwinding is arrested 

by the stop-work after four turns and a half, 
the spring will still be stretched one turn and 
a half when the watch stops. 

In most watches, in fact, this mechanism 
allows the barrel to make four revolutions 
on its arbor, and if it makes one turn in 
eight hours, as is often the case, the watch 
should run for 32 hours. 

The most modern stop-work is what is called the ‘ ‘ maltese 
cross.” It is composed of two pieces, the finger and the wheel* 
The latter being shaped like a maltese cross, gives it this name. 
The wheel is placed on the barrel, where it 
can turn freely, while the finger is placed 
on the arbor. The head of this piece gears 
in the notches of the cross, the rounded out 
teeth of which can successively slip around 
the circumference of the finger. 

On winding the watch, one turns the 
barrel arbor ; the finger participates then 
in this movement and pushes, at each turn, 
a tooth of the wheel until the moment when the shoulder of the 
finger comes into contact with the full tooth of the wheel ; the 
movement is then stopped, and the watch is wound (Fig. 24). 

During the running of the watch, the 
finger is stationary and the wheel, turning 
with the barrel, at each turn presents one 
of its openings in front of the end of the 
finger, which forces it to make a fraction of 
a revolution on its axis (Fig. 25). After 
the four revolutions of the barrel, the other shoulder of the finger 
comes again in contact with the full tooth of the wheel, and the 
watch is stopped (Fig. 26). 

128. Geometrical Construction of the Maltese Cross Stop-Work. 

In order to construct, graphically, this stop-work, we will suppose 
* These pieces arc commonly known ns the male and female.”! kanslator. 












8 o 


Lessons in Horology. 


that the distance O O' between the centers of the barrel and of 
the maltese cross wheel is known (Fig. 27). We divide this distance 
into five equal parts, and from the center O , with a radius equal to 
three of these divisions, one describes a circumference, which is 
repeated a second time from the point O' as center. From this 
last center we further describe a new circle passing through the 
center of the barrel, and we divide this last circumference into five 



equal parts. In order to get the end of the finger in one of the 
openings of the maltese cross, the division is commenced at the 
point 1'; for other cases we will commence to divide at the point O 
or any other intermediate point. 

The circumference of the finger is described with a radius 
equal to half of the distance O O '; from the points 1', 2', 3' and 5' we 
will trace with the same radius the arcs A C, A' C', etc., the arms 
of the cross. The intersections C A', etc., will, consequently, 
determine the size of the openings, the straight sides of which 
are drawn parallel to each other and at equal distances from the 
center O'. 
















Maintaining or Motive Forces. 


81 


The essential conditions to be fulfilled in making the end of 
the finger are the solidity of the piece and the free action of the 
mechanism. We prefer to represent it by means of two arcs of 
circles : one m n , whose length equals a semi-circumference, and 
the other 71 k , w’hose center is found almost on the point of the 
shoulder of the finger. 

Practically, the end of the finger k l should be slightly smaller 
than the corresponding opening of the other piece, that is to say, 
there should exist a certain play, to assure the free action of the 
mechanism ; this play will be easily obtained by taking off the sharp 
corners k and / of the finger. 

It is also to be recommended, in practice, to make the full 
tooth of the maltese cross with a radius O' D longer than the 
radius O' O of the cut-out teeth, in order to cause the stoppage 
a little before the line of centers. It is, moreover, necessary to 
slightly round off the corners A , C, A ', O', etc., of the teeth of 
the wheel. 


CHAPTER III. 


Wheel-Work. 

129. Purposes Of Wheel-Work. The wheels of a watch and of 
a clock have a double duty to fulfill : first, to transmit the move¬ 
ment arising from the motive power, from the first mobile down to 
the escapement ; second, to reckon the number of oscillations 
accomplished by the balance wheel in a given time, indicating this 
time by means of hands on a spaced dial. 

Since on the one hand the movement of the balance wheel is 
a rapid one and on the other the motive force should only be 
expended slowly, and, moreover, the wheels carrying the hands 
should make certain numbers of turns, according to given rela¬ 
tions, one understands that the wheel-work should be arranged in 
such a manner as to multiply, progressively, the speed of the first 
mobile. This is why we make the wheels gear into pinions, and 
the numbers of teeth of these different mobiles should be exactly 
determined. 

Let us further remark that in thus considerably increasing the 
speed, we diminish in the same proportion the force transmitted to 
the escape wheel. 

Calculations of Trains. 

130. Calculations of the Number of Turns. Let us determine, 
first, the relation which should exist between the number of turns 
made by the moving bodies of a gearing and their number of 
teeth. Knowing the number of teeth A in a wheel, and of leaves 
in the pinion <2, in which it gears, we have then to find the number 
of rotations accomplished, n\ by the pinion while the w’heel makes 
a number n. 

Let us suppose a wheel A of 96 teeth gearing in a pinion a of 
12 leaves : since each tooth of the wheel drives one leaf of the 
pinion, it is evident that, while the pinion makes one rotation, as 
many teeth of the wheel will have advanced as there are leaves in 
the pinion, and, consequently, the pinion will make as many turns 
as the number of times its leaves are contained in the teeth of the 
wheel ; one will have, then, 

„/ = ± = 96 = 8 . 
a 12 


82 




Wheel- Work, 


83 


If one wished to know the number of rotations completed by 
the pinion, while the wheel makes any number of them, n — 4, for 
instance, the number n’ will become n turns greater and one could place 

A v , 96 


(1) 


n' 


n 


a 


= 4 X 


12 


32 turns. 


The preceding equation can be presented under the form 

A 


(2) 


n' 

n 


a 


The members of turns made by the two mobiles are then 
inversely proportional to their 
numbers of teeth. 

131. Let us now consider 
a train of gearings formed of 
two wheels and two pinions 
(Fig. 28), the wheel with A 
teeth gearing in the pinion 
of a leaves, on the axis of 
which is fastened a wheel 
with B teeth gearing in a 
pinion with b leaves. 

While the wheel A com¬ 
pleted n rotations, B made n! and b completed n ". 

The number of turns that the pinion a makes while the 
wheel A makes a number 71, is expressed by the formula (1) 

, A 
n' — n —. 

a 

On the other hand, the number of turns that the pinion b makes 
while the wheel B makes n', should be 

n" — n f . 

0 

But since in this last formula 



n' = n 


A_ 

a 


one can write, after replacing n f by its value, 


or, further, 


( 3 ) 


n" — 71 


A 

a 


X 


( 4 ) Lr = 


B 
b 

A B 
71 a b 


— 71 


A B 
a b 


71 


// 


If, for example, a center wheel has 80 teeth, the pinion 
of the third wheel 10 leaves, the third wheel 75 teeth and the 










8 4 


Lessons in Horology. 


fourth pinion io leaves, one will find the number of turns that this 
last pinion should make while the center wheel makes i, by replacing 
the letters of the formula (3) by their numerical values, then 

n" = 1 X 80 = 60 turns. 

10 X 10 

The fourth pinion, then, makes 60 rotations while the center 
wheel makes one. Since the axis of the center wheel carries the 
minute hand and the axis of the fourth wheel carries the second 
hand, the movement will be executed, properly, according to the 
accepted division of time. 

If it were necessary to calculate the number of rotations of the 
fourth pinion, while the center wheel made 24, one would place in 
an analogous manner 

n" — 24 X 8 ° *- 75 = !44° turns. 

10 X 10 

132 . One could determine, in like manner, for any number of 
wheels and pinions, the relation of the numbers of turns of the last 
pinion to those of the first wheel. This relation is always equal to 
the quotient obtained by dividing the product of the numbers of 
teeth in the wheels by the product of the number of leaves in the 
pinions. One can then establish, in a general manner 

n" f _ A B CPE . 

^ n abode . 


Suppose, for example, we wish to determine the number of 
revolutions accomplished by an escape pinion while the barrel makes 
4, knowing that this barrel has 96 teeth, the center wheel also 
96, the third wheel 90, the fourth w r heel 80, and that all the pinions 
have 12 leaves, except that of the escape, which has 8. We would 
write the formula (5) under the form : 


( 6 ) 


n"" = 71 


A B CD 
abed ’ 


from whence, by replacing values, 

96 X 9 6 X 9° X 80 


n 


//// 


= 4 X 


12 X 12 x 12 x 8 


19200 turns. 


133. Calculation of the Number of Oscillations of the Balance. 

It is generally customary to indicate the number of oscillations which 
the balance wheel of a watch makes during one hour, that is, while 
the center wheel, which carries the minute hand on its axis, makes 
one turn. 












Wheel - Work. 


85 


134. We have already called attention to the fact that in most 
of the escapements the action of each tooth of the wheel corres¬ 
ponds to two oscillations of the balance (71). Knowing then the 
number of rotations which the escape wheel makes during one 
hour, one will easily calculate the number of oscillations which the 
balance executes during this same time, by multiplying the number 
of turns of the escape wheel by twice the number of its teeth, an 
operation which can be represented, designating the number of teeth 
of this last wheel by E and the number of oscillations by N, by 
the formula 

(7) N=2En"'. 


If an escape wheel with 15 teeth makes, for example, 600 turns 
while the center wheel makes 1, we will obtain the number of oscil¬ 
lations made by the balance by 

N = 2 X 15 X 600 = 18000 oscillations. 


135. If we designate by 


B f the number of teeth of the center wheel, 

C, “ “ “ “ “ “ third 

D, “ “ “ “ “ “ fourth “ 

b, “ “ “ leaves “ “ third pinion, 

c, ‘ ‘ “ “ “ “ “ fourth 

d, “ “ “ “ “ “ escape 


4 4 
< 4 


we should have, according to the formula (5) 


but since n — 1 


n'" — n 


B CD 
bed 


n'" = 


B CD 
bed 


Replacing n tff by this last value in the equation (7), we will 
obtain the general formula 


(8) N = 


B CD 2 E 
~b c d 1 


a formula which enables us to calculate the number of oscillations 
made by a balance wheel during one hour, knowing the numbers of 
teeth of the different mobiles. 

Suppose, for a numerical example, we desire to calculate the 
number of oscillations of a balance, knowing that 

B = 64 C — 60 D = 60 E — 15 
b ^ 8 c=8 d .= 6. 






86 


Lessons in Horology. 


The application of the formula (8) will give 

A . 64 X 60 X 60 X 2 X 15 Q Ml 4 .* 

N = o Ly-o 6 7 -- = 18000 oscillations. 

8X0X0 

136. Calculations of the Numbers of Teeth. Suppose now we 
wish to calculate the numbers of teeth in the wheels and of leaves 
in the pinions, the numbers of turns or of oscillations being known. 
This question, the reverse of the preceding one, can have several 
solutions ; in short, if one takes the equation 


n 


//// 


71 


A B CD 
abed, ’ 


in which the relation ~ alone may be known, and in which the 
unknown quantities may be A , B, C, D and a , b, c, d , one sees 
immediately that an unlimited number of values could satisfy this 
relation ; the equation is, in fact, indeterminate and affords as many 
unknown quantities as there are wheels and pinions. 

In order to determine them, one chooses arbitrarily the value 
of some of these unknown quantities, and, in order that the result 
will contain no fractions, one chooses for numbers of leaves in the 
pinions those employed in practical use. 

These numbers are, generally, 6, 7 and 8 for the escape pinions, 
8, 10 and 12 for the pinions of the third and fourth wheels, 10, 12 
and 14 for the pinions of the center wheels. 

The values a, b , c , d becoming, thus, known quantities, could 
be transposed to the first member of the equation, which will be 
written under the form 


7l"" 

- abed = AB CD. 

71 

In order to solve this equation, it will suffice, then, to resolve 
all the known numbers of the first member into their prime factors 
and to form these factors into as many groups as there are unknown 
quantities to be determined. 

Let us take a numerical example, and suppose that the relation 
in the above equation be 


7 l"" 

71 


— 4800. 


We will then have 


4800 


A B CD 
abed ’ 


let us choose the following numbers of leaves for the pinions : 
a = 10, b = 10, c = 10 and d — 7, 
we will place 

4800 X 10 X 10 X 10 X 7 = A B CD. 








Wheel - Work. 


87 


Resolving 4800 with its prime factors, we obtain 


4800 

2400 

1200 

600 

3 °° 

150 

75 

25 

5 

1 


2 

2 

2 

2 

2 

2 

3 
5 
5 


4800 = 2 6 X 3 X 5 2 

10=2X5 

and 7 is already a prime number. 

One will have then the total product : 

2 9 X 3 X 5 s X 7 = 4800 X 10 X 10 X 10 X 7 = A B CD , 

with the factors of which we can form the following 
groups: 


A = 

2 4 

X 

5 = 80 

or 

A = 2 2 

X 5 2 

= IOO 

B = 

2 4 

X 

5 =80 

< < 

B = 2 2 

X 3 x 7 

= 84 

C = 

3 

X 

5 2 = 75 

< < 

C= 2 4 

X 5 

= 80 

D = 

2 

X 

5 X 7 = 70 

i < 

D = 2 

X 5 2 

= 50 


etc., etc. 


By employing other numbers of leaves for the pinions, one can 
multiply the solutions to infinity. As proof, we could have 


4800 


80 X 80 X 75 X 70 too X 84 X 80 X 50 


10 X 10 X 10 X 7 10 X 10 X 10 X 7 

137. When the relation — is fractional, one factors the nume- 
rator and denominator separately, then cancels the common factors. 
In a case where this elimination could not be effected, the problem 
would become impossible with the number of leaves chosen, and it 
would be necessary to replace them by others. 

For example, let the formula be 

360 


n 

n 


// 


It is impossible to solve this case with two pinions of 10 leaves, 


for in 


3 60 X i Q X iQ _ 2 5 X 3 2 X 5 3 
7 7 


there exists no factor 7 in the numerator which could serve to elimi¬ 
nate that of the denominator. On the other hand, if one chooses 
the two numbers 14 and 10, one will have 

360 X 14 X 10 _ 2 5 X 3 2 X 5 2 X 7 
7 “ 7 

and will be able to cancel the factor 7. One forms, then, two 
groups with the figures which remain, and obtains, for example, 

A — 2 4 X 5 — 80 











88 


Lessons in Horology. 


As proof, one will have correctly, 

80 X 90 _ 360 

10 X 14 ~ 7 

138 . Let the question be, now, to determine the numbers of 
teeth in the wheels of a watch, whose balance should make 16200 
oscillations per hour. The formula (8) gives 

, B CD 2 E 

16200 — — . —,— ; 

bed 

let us choose for the pinions the following numbers of leaves : 

b — 12, c — 10, d = 8; 

since the number of teeth in the escape wheel varies only within 
very narrow limits, we can further replace the letter E by the 
figure 15, for example. This number is, in fact, that which is very 
generally used for watches of medium size. One will then have : 


16200 


D CDX 2 X 15 
12X10X8 


or, on transposing the known terms, 


16200 x 12 X IQ X 8 __ B D 

2 X 15 


Further simplification gives 

16200 X 2 X4 X 4 — BCD. 
Resolving into prime factors, one obtains 


2 8 X 3 4 X 5 2 = B CD, 

with which one can form the following groups : 


B = 2 X 3 2 X 5 = 90 

C — 2 4 X 5 =80 

D — 2 3 X 3 2 = 72. 


The verification of the operation should give 


16200 


90 X 80 X 7 2 X 2 x 15 

12 X 10 X 8 


We will occupy ourselves, in the problems which follow, with 
the numbers of teeth to be given to the barrel and to the pinion of 
the center wheel. 

139. The number of oscillations of the balance varied greatly 
in the earliest watches ; this figure was governed by no fixed rule 
and would vary between 17000 and 18000. As these watches had 









TV heel - Work. 


89 

no second hands, this number had only a relative importance 
within these limits. 

In our modern timepieces, especially in watches above 12 
lines (27 mm.), five oscillations per second, which would be 18000 
per hour, are generally adopted. 

In smaller pieces, this figure is increased to six oscillations 
per second, being 21600 per hour, with the object of diminishing 
the influence of jars in carrying, always very perceptible on the 
small balance with which these watches are supplied, 

A great many of the English watches beat 16200 oscillations, 
being 4^ per second. 

Marine chronometers beat four oscillations per second, or 
14400 per hour. 

140 . The problems which follow are the applications of the 
preceding theories and will aid in the better understanding of these 
various questions. We especially insist on the constant use of the 
formulas, and urge the pupils to accustom themselves to solve these 
questions by applying to each case the equation which suits it. 
The algebraic way is a sure guide which leads always to a correct 
solution and to exact results. In the following exercises we give 
numerous examples of the reliability of calculation which results 
from the use of the simple formulas which we have just established. 


Problems Relative to the Preceding Questions. 

141 . A barrel of 80 teeth gears in a center pinion with 10 leaves; 
how many turns will this pinion make while the barrel makes 1 ? 
Solution : We have the formula (1) which gives 


7 l' = 71 


A 

a 


= 1 X 


80 

10 


= 8 . 


The pinion executes 8 turns while the barrel makes 1 ; one revo¬ 
lution of the barrel has then, in this case, a duration of 8 hours. 

142 . How 771 any turns will this same pinion make while the 
barrel makes 4. f 

Solution : The formula (1) further gives 


, A 80 

„ _= 4 X - ■= 32. 


While the barrel makes 4 turns the center pinion makes 32. 
Since the stop-works are placed in such a manner that the barrel 



90 


Lessons in Horology. 


can execute exactly four rotations on its axis (127), the watch will, 
therefore, run for 32 hours. 

143. A center wheel with 64 teeth gears in a third wheel pinion 
with 8 leaves ; the third wheel with 60 teeth gears in a pmion with 
8 leaves also. How many turns will this last pmion make during 
one turn of the center wheel ? 

Solution : The formula (3) 


71 


// 


71 


gives us, after substituting, 


A B 
a b 


n 


7 / 


I X 


64 X 60 

‘ 8 x 8~ 


= 60. 


The fourth pinion will make 60 rotations during one revolution 
of the center wheel, therefore, during one hour. 

144. What is the member of rotations which a7i escape pinion 
with y leaves will make during 12 hours , knowing that the center 
wheel with So teeth gears in a pinio7i of the third wheel with 10 
leaves , the wheel of which with yy teeth gears in the fourth pinion 
with 10 leaves ; the fourth wheel having yo teeth f 

Solution : We will use the general formula for a train of three 
gearings: 


71 = 71 


ABC 
a b c ’ 


from whence, after substituting values, 

80 X 75 X 70 


71 


777 


= 12 X 


7200 turns. 


10 X 10 X 7 

145. What is the number of hums exeaited by an escape pinion 
during j turns of the barrel , the wheel-work having the following 
teeth-ranges : 

Barrel .... 96 teeth , Ce7iter pinion 12 leaves , 

Center wheel 90 ‘ ‘ Third “ 12 “ 

Third “ 80 “ Fourth “ 10 

Fourth “ y2 “ Escape “ 8 

Solution : Using the formula, we have 


< < 


< < 


7l"" = 71 


A B C D 
abed ’ 


or 


71 


77 // — 


= 3 X - 


96 X 9° X 80 X 72 


12 X 12 X 10 X 8 


= 12960 turns. 


146 . Suppose we wish to calculate the number of oscillations 
which a balance makes during one hour; the center wheel having 
64 teeth , the third wheel 60 , the fourth wheel 56, the escape 








Wheel - Work 


9i 


wheel 15 ; the pinions of the third and fourth wheels each 8 leaves 
and that of the escapement 7. 

Solution : The formula (8) gives 


from whence 


N = 


B C D 2 E 
bed * 


AT _ 6 4 x 60 x 56 X 2 X 15 
8X8X7 


— 14400 oscillations. 


147 . What should be the number of teeth in a fourth wheel 
gearing in an escape pinion , knowing that the pinion should make 
10 turns while the wheel makes 1 ? 

Solution : The equation (2) 

nf = A _ 
n a 


gives, after replacing n' and n by their values, 

10 A 


this equation with two unknown quantities, A and a , is indefinite ; 
several solutions can, therefore, satisfy its demands. Replacing a 

successively by the numbers 6, 7, 8, 10., we find for A the 

corresponding values, 

60, 70, 80, 100.. 

because 

10 60 70 80 100 

1 6 7 8 10 


One obtains, then, the number of teeth in the wheel by multi¬ 
plying the number of leaves chosen, by the number of rotations 
which the pinion should make. This is always practicable when 
the number of turns is a whole number. 

If, in place of choosing the number of leaves in the pinion, one 
takes the number of teeth in the wheel, the result may easily become 
fractional. 

The equation (2) can be written by making n = 1 

A 

a = — 7-. 
n ' 

Let A = 66 , we would have for the preceding case, 

66 

a = —- 

10 

a solution impossible to carry out. 


= 6 |, 












92 


Lessons in Horology. 


It is then preferable to choose, always, the number of leaves in 
the pinion, and to determine from these the numbers of teeth in 
the wheels. 

148 . To determine the member of teeth in a third wheel and the 
number of leaves in a fourth pinion combined in such a manner that 
the pinion makes 15 rotations while the wheel makes 2. 

Solution : One will use (2) 

n ' __ A 
n a 

and, after substituting 

£5 _ ± 

2 a 

Since 15 and 2 are prime to each other and it is, therefore, 
impossible to simplify their relation, it is necessary, in order to 
avoid fractional numbers, that A be a multiple of 15 and a a 
multiple of 2 ; thus one could have 

25 = = 21 = J 2 etc. 

2 6 IO 12 

The numbers 45, 75, 90 will, therefore, be suitable for the wheel A, 
and 6, 10, 12 for the pinion a. 

149 . We wish to k?iow the number of teeth in a barrel and the 
7 iumber of leaves in the center pinion in which it gears , so that the 
watch may mm 8 days with 4 turns of the barrel. 

Solution : From the equation (2) we find the value 

A n ' 

A — — a , 

71 

in which n f = & X 24= 192 and n = 4 ; then, 

A = —-a = 48 a . 

4 

Replacing a by the numbers 6, 7, 8 and 10, successively, one 
finds that 

A = 48 X 6 =288 

A = 48 X 7 = 336 

A — 48 X 8 = 384 

A = 48 X 10 = 480. 

These solutions have the disadvantage of giving too great a 
number of teeth to the barrel, for even on choosing for a , a pinion 
of 6 leaves, one obtains, still, a barrel with 288 teeth. 

In order to avoid this inconvenience, one sometimes has 
recourse to an intermediate pinion between the barrel and the 




Wheel- Work. 


93 


center wheel ; the barrel would gear in this pinion and the wheel 
mounted on the axis of this pinion should gear in that of the center 
wheel. The difficulty is thus changed and becomes that of finding 
a place in the watch in which to put another mobile and of increas¬ 
ing the motive power a very appreciable quantity. 

In order to solve the problem thus arising, we will make use 
of formula (4) 


and we will have 


n" = A B 
n a b 


192 A B 

4 a b 


— 48. 


Choosing for a 12 leaves and for £ 10, one has 

A B — 48 X 12 X 10. 

Resolving the two members of the equation into their prime 
factors, one will obtain 

2 7 X 3 2 X 5 = A B, 


with w'hich one could form the two groups 

A = 2 3 X 3 2 = 72 
B — 2 4 X 5 = 80, 

or else 

A = 2 5 X 3 —96 

B = 2 2 X 3 X 5 = 60. 

As proof, one should find that 

8 _ 72 X 80 _ 96 X 60 
12 X 10 12 X io* 


150 . Suppose we wish to determine the nu7nbe? r s of teeth a7id 
of leaves in the wheels and pinio 7 is formmg the dial wheels. 
Descriptio7i of this mecha7iism. 

The dial wheels are the mechanism whose object is to secure 
the movement of the hour hand. Since the center wheel makes one 
turn per hour, one fixes on the prolongation of its axis, under the 
dial, a second pinion, called the ca7mo7i pi7iio7i. This adjustment is 
made in such a manner that this cannon pinion participates in the 
movement of the center wheel during the ordinary running of the 
watch, although it is possible to give it a separate movement when 
one wishes to set the hands. Thus the center wheel, the cannon 
pinion and the minute hand have a common movement and make 
one turn per hour. The cannon pinion a gears in the 77 imute wheel 







94 


Lessons in Horology. 


A (Fig. 29), which carries a pinion b gearing in the hour wheel B. 
This last wheel, usually placed on the cannon pinion, around which 
it can turn freely, is the one which carries the hour hand. The 
hour wheel should, therefore, make one turn in 12 hours ; other¬ 
wise expressed, the cannon pinion should complete 12 rotations 
while the hour wheel makes one. 


rn 111 mTiiiTrmn 


t 

A 

in minimum 


Hour Hand 


^Minute Hand 

Fig. 2 J) 

In the equation (4) one replaces n" by 12 and n by 1, the 
unknown quantities are then A B and a b ; one, therefore, has 

12 A B 

— — -T— = 12, 

i a b 

and substituting for a and b the numbers 12 and 10, one will have 

12 X 12 X 10 = A B. 

Resolving into prime factors, it becomes 

25 X 3 2 X 5 = A B, 

which we can group in the following manner : 

A = 2 2 X 3 2 = 36 
B = 2 3 X 5 =40. 

_ 36 X 40 

12 x 10' 



As proof, one has 


12 





























































Wheel - Work. 


95 


These figures, 36 for the minute wheel and 40 for the hour 

wheel, are very often employed in practice ; one then gives 12 

leaves to the cannon pinion and 10 to the minute wheel pinion. 
Evidently other groups can be formed, such as these : 

A = 2 3 X 3 =24 

B = 2 2 X 3 X 5 = 60, 

or 

A = 2 5 = 32 

B = 3 2 x 5 = 45 , 

or, again, 

A = 2 X 3 X 5 = 3° 

B = 2 4 X 3 —- 48. 


The verification always gives : 

24 X 60 _ 32 X 45 _ 30 X 48 _ i2 
12 X 10 12 X 10 12 X 10 

In small watches or low-priced ones, a cannon pinion of 10 
leaves and a minute-wheel pinion of 8 leaves are often used ; this 
gives for the wheels : 

12 X 10 X 8 — A B 

and 

2 6 X 3 X 5 — A B. 

The two groups ordinarily employed are : 

X = 2 X 3 X 5 = 3° 

B = 2 5 = 32. 

Sometimes, also, a cannon pinion of 14 leaves is used and a 
minute-wheel pinion of 8 leaves ; it then becomes 

12 X 14 X 8 = A B 
2 6 X 3 X 7 = A B, 

from which one can make 

A — 2 2 X 7 = 28 
B — 2 4 x 3 = 4 8 - 

These last two cases always give 

3 ° X 3 2 28 X 48 _ 

10 X 8 14 X 8 

151 . If it were desired to make the dial wheels of a watch 
whose dial was divided into 24 hours, the question would not be 
any more complex, since one would only have to solve the equa¬ 
tion 24 a b = A B. 







96 


Lessons in Horology. 


Take, for example, a — 12 and b = 10, one would have 


and 

from whence 

Proof 


24 X 12 X 10 = A B 

2 6 X 3 2 X 5 = A B y 

A — 2 4 X 3 =48 

B — 2 2 X 3 X 5 = 6o - 


48 X 60 
12 X i° 


24. 


152 . If the same dial ought to show, by means of two pairs of 
hands of different color or shape, the division of time into 24 hours 
and the division into 12 hours (Fig. 30), it would be easy to use 



the same cannon pinion and the same minute wheel for both sets 
of wheels ; one need only add a second pinion, c, fastened on top of 
the first, and gearing in a second hour wheel, C , loosely fitted on 
the first wheel. 

Admitting for the first train 

Cannon pinion 12 leaves, Minute wheel 36 teeth, 

Minute pinion 10 leaves, Hour wheel 40 teeth, 

one should have for the first train 

3 6 X C 

12 X c 


24 = 





Wheel- Work. 


97 


Choosing for c 6 leaves, one will have 


from whence 


C = 


one will have correctly 


2 A = 36 X C 
12 X 6’ 

24 X 12 x 6 
36 

24 = 36 .X 48 

4 12 X 6 


48; 


The two minute hands are fastened on the axis of the center 
wheel, since in both cases they should execute one turn in an 
hour ; their angle of divergence once being determined, will remain 
permanent. 

153 . Calculation of the numbers of teeth in the wheels of an 
astronomical clock {seconds regulator ) which should run 33 days , 
the weight having a drop of 830 mm. The cord unwinds from a 
cylinder whose raduis is 13 mm ., in which is included half the 
thickness of the cord. This cord is supposed to run through a 
pulley. 

Solution : Since the cord runs through a pulley, it unwinds 
from the cylinder a length equal to twice the descent of the weight, 
therefore, 1660 mm. On dividing this length by the circumference 
of the cylinder, 2 w r, we will obtain the number of turns executed 
by this cylinder during the descent of the weight; therefore, 


1660 

2 X 3.1416 X 15 


= 17.6 


. turns. 


The cylinder makes, then, 17.6 turns, while the weight 
descends 830 mm.; or, according to the data, during 33 days, 
or, again, during 24 X 33 = 792 hours. 

One turn of the cylinder will then be effected in 


~7g = 45 hours. 
17.0 


Consequently, while the wheel fixed on the arbor of the 
cylinder (fusee wheel) makes one turn, the wheel carrying the 
minute hand must execute 45. 

One sees at once that, in order to avoid having too great a 
number of teeth, one should introduce an intermediate pinion a 
and wheel B between the fusee and the center wheel (Fig. 31). 







9 8 


Lessons in Horology. 


In order to determine the numbers of teeth in the fusee and in 
the intermediate wheel, as well as for the pinions a and b ) one 
would employ the equation (3) 



in which n' f — 45 and n = 1. If we should choose pinions of 
38 and 16 leaves, we will place 

45 X i 3 X 16 = A B. 

The first member separated into prime factors gives 

2 5 X3 4 X5 = ^^ 









Wheel - Work. 


99 


\ 


which can be grouped in the following manner : 

A = 2 2 X 3 s = 108 
B = 2 3 X 3 X 5 = 120 

We would have, correctly, 

108 X 120 


To determine the numbers of teeth suitable for the other 
mobiles, let us note, first, that since the pendulum of this regulator 
should beat one oscillation per second, an escape wheel with 30 
teeth should execute one turn in a minute (71). One can then 
fasten the second hand on the prolongation of the axis of its 
pinion d. The escape wheel executes then 60 turns, while the 
center wheel makes 1, and one will have, on employing pinions 
of 12 and 10 leaves, 60 X 12 X 10 = C D, 

° r . 2 5 X 3 2 X 5 2 = C D, 

which can give „ w „ w 

s C = 2 X 3 2 X 5 — 9 ° 

D = 2 4 X 5 = 80. 


As proof, one will have correctly 

* 90 X 80 

60 = —— -- 

12 x 10 


154. if ; in place of running jj days , one desired a clock 
running ij months , what should be the numbers of teeth in the 
wheel-work , with the same data as that in the preceding problem ? 

Solution : Thirteen months calculated at the rate of 30 days is 
equal to 390 days or 9360 hours. One places 


n 


/// 


71 


ABC 
a b c ’ 


for one sees that, in order to avoid having wheels with too many 
teeth, a second intermediate wheel must be introduced between the 
fusee and the center wheel. One will then have 


9360 _ ABC 

17.6 a b c 

that is to say, while the fusee wheel makes 17.6 turns the center 
wheel should make 9360. 

Since the numerical expression 

9360 

17.6 

cannot be employed because of the fraction in the denominator, we 
will transform it by means of the following operation into an 










IOO 


Lessons in Horology. 


equivalent fraction, having as denominator a whole prime number : 


9360 

17.6 


531it ; 


multiplying this quotient by 11, one obtains the whole number 
5850, then 

9360 __ 5850 
17.6 11 

One will have, consequently, 

abc = A B C. 

11 

In choosing the numbers of leaves for the pinions, care must 
be taken that 11 is found as factor in one of these numbers, in 
order to be able to eliminate the denominator. Let us take, then, 
the figures 22, 16 and 14 ; we will have 


-^2 x 22 X 16 X 14 = A B C 

and 

X 3 2 X 5 2 X 7 X 11 X 13 —ABC. 

11 


After cancellation, one could form the following groups : 

A — 2 6 X 3 = 192 

B = 2 X 5 X 13 = 130 

C = 3 X 5 X 7 — 105. 

If one found these values too great, one could choose pinions 
with fewer numbers; for example, 10, 11 and 12 leaves, and one 
would have, in an analogous manner, 

A = 2 3 X 3 X 5 = 120 

B = 2 X 3 X 13 = 78 

C = 3 X 5 2 = 75 * 

For both cases we would have correctly, 


192 X 130 X 105 _ 1 20 X 78 X 75 = 5850 
22 X X 14 12 x 11 X 10 11 

For the other wheels of the train, the case is the same as in 
the preceding example. 

By introducing a third intermediate wheel, one could succeed 
in making such a clock run for 10 years, but there exist practical 
disadvantages which make this combination seldom used. 

155 . How to determine the number of teeth in a third wheel 
which has been lost y knowing that the balance should beat 18,000 
oscillations per hour and knowing the numbers of teeth in the other 
wheels and pinions ? 










Wheel - Work. 


IOI 


Solution : Let us call x the unknown number and let the 

Center wheel have 80 teeth, Third wheel pinion io leaves 

Third “ “ x “ Fourth “ “ io “ 

Fourth “ “ 70 “ Escape “ “ 7 “ 

Escape “ “ 15 “ 

The formula (8) admits of placing 

,8ooo = 80 X * X 7 ° X » X 15 
10 X 10 X 7 

or, simplifying, 

18000 = 240 x , 

and 

„ 18000 . .. 

A = -- — 75 teeth. 

240 

The lost third wheel, therefore, had 75 teeth. 

156 . If , in the preceding problem , the last mobile had been the 
third-wheel pinion , how would the eqiiation be solved? 

Solution : We would have in an analogous manner : 


or 


and 


18000 = 


80 X 75 X 70 X 2 X 15 
x X 10 X 7 


_ 180000 

18000 = - 

x 


X 


180000 

18000 


— 10 leaves. 


157 . Still using the preceding data , let us suppose that the 
pinion and the escape wheel were both lost, and let us propose to 
detemnine their teeth ranges. 

Solution : We will have, in this case, two unknown quantities, 
which we will designate by ;tr andjy; the equation (8) will be written 


from whence 


and 


On simplifying, 


18000 = 


80 X 7 5 X 70 X 2 x 
10 X 10 X y 


18000 = 


8400 x 

y 


18000 X 

8400 y 


15 _ x 
7 y ' 


The wheel, then, should have 15 teeth and the pinion 7 leaves. 












102 


Lessons in Horology. 


158. In the last problem , we arrived immediately at the real 
numbers; this does not always happen. Let it be desired , as a 
second example , to find the numbers of teeth and of leaves in a 
center wheel and a third pinion which have been lost: 

Solution : One has 


18000 = 

from whence one obtains 
and 

the simplification gives 


X 75 X 70 X 2 X 15 

y X 10 x 7 


x 


18000 — — 2250, 

y 


18000 _ X 

~ y 


2250 


y 


= 8 . 


This result shows us that the center wheel should have eight 
times as many teeth as the pinion has leaves. On replacing succes¬ 
sively^ by 6, 7, 8, 10 and 12 leaves, one will obtain the following 
solutions : 


For y — 


< < 


i < 


y = 
y = 
y = 
y = 


6 one has x — 6X8 = 48 teeth 

7 “ ^ = 7 x 8 = 56 “ 

8 “ x = 8 X 8 = 64 “ 

10 “ x = 10 X 8 = 80 “ 

12 “ = 12 X 8 = 96 “ 


Several solutions can, therefore, satisfy the demand, and the one 
which suits best must be chosen ; it is evident, here, that with rela¬ 
tion to the numbers of teeth in the other mobiles, a center wheel 
with 80 teeth and a pinion with 10 leaves are perfectly admissible. 

159. Indicator of the Spring’s Development in Fusee Timepieces. 
Marine chronometers and a great many fusee watches carry an 
auxiliary hand placed on the dial, and with its center on a straight 
line between the point of XII o’clock and the middle of the center 
wheel. The object of this hand is to indicate, on a small dial, the 
number of hours which the chronometer has run since it was last 
wound. It gives notice, thus, of the proper time for rewinding 
the chronometer, an operation which then brings the hand back 
to zero. 

This mechanism, easy to establish in fusee watches, becomes 
difficult to introduce in other kinds. Let us take up the first for 
the present. 





Wheel- Work. 


103 


When one winds one of these timepieces, one causes the axis 
of the fusee to turn, which, once the spring is wound, takes a 
movement in the opposite direction. If, therefore, one places on 
this axis a pinion communicating its movement to the wheel on 
which the small hand is fastened, the mechanism will be complete. 

The indicating dial is arranged in such a way that the figure XII 
may be wholly preserved ; the hand cannot, therefore, make a com¬ 
plete turn. Let us suppose that while the watch runs 56 hours 
with 8 turns of the fusee, this hand may make seven-eighths of a 
turn and that there thus remains one-eighth of a turn, which is 
taken up by the lower part of the figure XII (Fig. 32). 



t ig. 3I£ 


The question now, is to determine the numbers of teeth in the 
wheel carrying the hand and in the pinion fastened on the axis of 
the fusee, in order to produce a movement conforming to the 
given data. 

While the fusee executes n' = 8 turns, the hand should make 
n= \ \ therefore, (2) 

n' _ A _8_ _ 64 = A_ 
n a | 7 a 

The number of teeth in the wheel should, therefore, be a mul¬ 
tiple of 64 and the number of leaves in the pinion a multiple of 7. 

One can have for 

a = 7, A = 64 

a — 14, A — 128, etc. 

When the chronometer is running, the fusee is animated with 
a movement to the left ; the hand, therefore, turns to the right, in 
the same direction as the other hands. 







104 


Lessons in Horology. 


It may happen that, owing to the arrangement of the calibre 
of the watch, one could not make the pinion gear directly in the 
wheel ; it would be necessary in this case to place a second wheel 
gearing on one side in the fusee pinion, and on the other in the 
wheel carrying the hand. This last mobile will then take a move¬ 
ment to the left, and if one wished to avoid that, it would be neces¬ 
sary to arrange two intermediate wheels between the wheel A and 
the pinion a. 

Let us note that the number of teeth in this or in these inter¬ 
mediate wheels does not modify the relation existing between the 
movement of the wheel A and that of the pinion a. Designating 
by B the number of teeth in the intermediate wheel, one has 

n' _ A_B _ A 
71 a B a 

160 . Simple Calendar Watches. By this name we designate a 
certain class of watches having an accessory mechanism by means 
of which is shown, on the dial of the watch, the date, the day of 
the week and the name of the month. These indications are made 
by means of hands fixed on the axes of toothed wheels, performing 
their revolution in the length of time desired ; that is to say, the 
hand indicating the date jumps each day at midnight, as does also the 
one indicating the day of the week. These movements are obtained 
by means of a wheel making one rotation in twenty-four hours. 
The dial showing the date is divided into thirty-one parts ; the hand 
jumps, therefore, for months of thirty-one days from the figure 31 
to the figure 1. If the month has only twenty-eight, twenty-nine 
or thirty days, the hand should be set to the figure 1 by hand. 
This is inconvenient and is overcome in perpetual calendars, but 
such mechanisms are, therefore, more complicated. 

Calender watches are often made to indicate also the phases of 
the moon. An opening is made in the dial for this purpose, across 
which pass successively representations of the various appearances 
which the moon presents. The movement of this satellite is not 
continuous, as in nature, but intermittent ; it is produced each day, 
as are the other actions of the calendar. 

Let us examine now in what manner the different movements 
which we have just mentioned, are effected (Fig. 33). 

The most simple construction consists in placing a wheel of 
30 teeth on the hour wheel ; this wheel gears in two other wheels, 



Wheel- Work. 


105 


B and C, of 60 teeth each. These last mobiles execute, therefore, 
one revolution in 24 hours. The wheel B carries a pin m perpen¬ 
dicular to its plane, which comes into gear at each rotation with the 
teeth of a “star” of 31 teeth carrying the calendar hand. This 



last wheel is kept in place by a jump-spring. During the day this 
same pin makes the phase wheel L jump in the same manner. This 
wheel, also star-shaped, contains 59 teeth ; it is in like manner 
kept in place by a jumper ; on its surface are represented two 
moons diametrically opposed. When one of the faces disappears 
behind the dial at the time of the new moon, the edge of the fol¬ 
lowing is on the point of appearing in the shape of a slight crescent. 

A synodical revolution of the moon (interval comprised be¬ 
tween two consecutive full moons) is effected in very nearly 29 days 
and a half; * two lunations require about 59 days. This is the 
reason why 59 teeth are given to the phase wheel. 


♦ Exactly, 29 days, 12 hours, 44 minutes. 















io6 


Lessons in Horology. 


The second wheel C also carries a pin, intended to make the 
wheel S y with 7 teeth, jump ; this wheel carries the hand indicating 
the days of the week. 

The movement of the hand which indicates the month, is most 
generally effected by setting it by means of an exterior push-piece. 
This hand is carried by a star wheel with 12 teeth, which is kept in 
place during one month by a jump-spring similar to those of the 
other stars. 

It does not require, therefore, much calculation in order to de¬ 
termine the toothing of these different wheels. 

161 . Suppose it be desired to determine the numbers of teeth 
and leaves of the wheel-work in a decimal watch , desiring to pre¬ 
serve to the balance wheel of this mechanism the same duration of 
oscillation as in that of an ordinary watch. 

Solution : A decimal watch is an instrument dividing the length 
of a day into twenty parts, in place of twenty-four ; therefore, the 
interval included between a midnight and a midday, or a midday 
and the following midnight, into ten equal parts. Each of these 
“hours” is divided into 100 “minutes.” 

Let us further make this a condition : this watch should run 
just as long as an ordinary watch (32 duodecimal hours). An 
ordinary watch, furnished with the customary stop-works, can 
run 1 y$ days while its barrel makes four turns ; this barrel will 
execute, therefore, 3 turns in a day. In a decimal watch, the 
center wheel should, accordingly, make 20 rotations while the 
barrel makes 3. 

While this barrel makes 1 turn, the center wheel will make 


One will, therefore, have 


20 

T 


= 6% turns. 



Choosing a pinion a with 12 leaves, a multiple of the denomina¬ 
tor of the fraction, it will become 

6 % X 12 == A, 

from whence 

A — 80. 

The barrel would, therefore, have 80 teeth and the center 
pinion 12 leaves (Fig. 34). 



Wheel- Work. 


107 

The fourth wheel should then execute 100 turns while the 
center wheel made 1 ; we will, therefore, have 

B C 

100 — — 
a b 

Choosing pinions of 8 leaves, we have 

100 X 8 X 8 — B C. 

Reducing to prime factors, one obtains afterwards 

2 8 X 5 2 = B C, 

one could form the two groups 

2 4 X 5 = 80 
2 4 X 5 = 80, 

The center and the third wheel should, therefore, each have 80 
teeth and should gear in pinions with 8 leaves. There now remains 
for us to determine the numbers of teeth in the fourth and escape 



wheels, as well as the number of leaves in the escape pinion ; 
these numbers have to fulfill the condition declared, not to alter 
the duration of the oscillations of the balance. 

Let us determine, in the first place, the number of oscillations 
which the balance would execute during one turn of the fourth 
wheel. In one day this number is equal to 24 times 18,000 ; for 
one turn of the center wheel it should be 20 times less, and for 
one turn of the fourth wheel still 100 times less, which gives 


24 X 18000 
20 X 100 


— 216 oscillations. 


Let us admit, as is the custom, an escape wheel with 15 teeth. 
The number of turns which this wheel should execute while the 





io8 


Lessons in Horology. 


fourth wheel makes one, will be obtained by dividing 216 oscilla¬ 
tions by twice the number of teeth in the escape wheel, therefore, 


One, therefore, places 


216 

2X15 


= 75 turns. 



D 
d ’ 


and, on choosing for the number of leaves a multiple of 5, 10, for 

example, one will have 

i*i • 74 X 10 = D, 

which gives 

D — 72 teeth. 


The fourth wheel could have 72 teeth and the escape pinion 10 leaves. 

162 . In the problem with which we have just dealt, the second 
hand will not divide the minute into 100 parts, since it will make 
216 little jumps during one revolution. We could still propose to 
divide the minute into 100 equal parts, by abandoning the condition 
stipulated in the first problem, of keeping for the balance the same 
duration of oscillations ; in place, therefore, of making it execute 
216 oscillations, let us imagine it as making 200 of them. 

With an escape wheel of 15 teeth, one arrives at 


200 

2X15 


= 6% turns 


executed by the escape wheel while the fourth wheel made 1. 
Choosing a pinion of 9 leaves, one has 

6 2 A X 9 = 60 teeth 

for the fourth wheel. 

163 . We have still to make the calculation of the wheel-work 
for the dial wheels. This problem can have two forms : the one 
in which the hour hand should execute 1 turn a day, and the one 
in which it should make 2. 

In the second case, the minute hand will make 10 turns while 
the hour hand makes 1 ; one will, therefore, write 


10 = 


A B 
a b 


with pinions of 12 leaves each, one will have 


10 X 12 X 12 = A B \ 
on reducing into prime factors, 

From whence 2 ^ ^ ^ $ A B. 

A = 2 3 X 5 =40 teeth 
B = 2 2 X 3 2 = 36 “ 






Wheel- Work. 


109 


If the hour hand should only make one turn a day, one then has 

A B 


20 


Taking b 


a b 


10 and a — 8 : 

20 X 10 X 8 = A B t 


or 


X 5 J 


A B. 


One could then form the two groups 


A = 

B = 


X 

X 


5 = 
5 = 


40 

40. 


The minute and hour wheels would each have 40 teeth in this case. 

164. Calculation of the Numbers comprising the Teeth-ranges 
of the Wheels of a Watch with Independent Second Hand. These 
watches, which were 
constructed in consid¬ 
erable numbers some 
years ago, generally 
contained two distinct 
trains. In this system 
a special hand is placed 
at the center of the 
dial and makes one 
jump only per second ; 
it can be arrested for 
an indefinite time, then 
started again at will, 
without stopping the 
watch. The office of 
this second train is to 
drive this independent 
second hand. The 
principle of the me¬ 
chanism is, therefore, 
to release, at each sec¬ 
ond, the train which 
brings the hand into 
action. For this pur¬ 
pose the last pinion of 
the second train car¬ 
ries on its axis an arm 




I IO 


Lessons in Horology. 


called the “whip,” gearing either directly in the escape pinion of 
the first train or in a “star” adjusted on the axis of this latter 

(Fig 35) - 

While the whip is in contact with a leaf of the escape pinion, 
it has a slightly-pronounced angular movement, scarcely percept¬ 
ible on the second hand. But when the leaf of the pinion has 
advanced up to a certain point, the whip becomes free and rapidly 
makes almost a complete turn and again comes in contact with the 
pinion on the next leaf. At each turn of the w'hip the second 
hand should advance one division on the dial. At each second, 
therefore, a leaf of the pinion or a tooth of the star must present 
itself to receive the whip. 

A lever or a cylinder escapement, whose wheel advances at 
each vibration of the balance, half the space which separates two 
consecutive teeth, can serve for this purpose, if the number of 
oscillations is 18,000 per hour, therefore, 5 per second. In effect, 
a wheel with 15 teeth produces 30 oscillations and requires, therefore, 

70 

—- = 6 seconds 

5 

to make one turn. 

If one causes the whip to gear directly into the pinion, the 
latter should have 6 leaves ; if not, it would be necessary to fasten 
a star with 6 teeth, on its axis, into which the whip should be made 
to gear. The movement will then be effected according to the 
requirements. 

There is a remark to be made about watches provided with the 
detent or duplex escapements. 

During the vibration in which the wheel gives the impulse to 
the balance, this wheel advances an angle equal to that which 
separates two consecutive teeth, and during the succeeding oscilla¬ 
tion it remains at rest. Owing to this fact each tooth still pro¬ 
duces two oscillations ; but we cannot then allow the balance to 
make 18,000 oscillations, because the whip should become free at 
the end of every five vibrations and, the figure 5 being an odd 
number, there would be found, every two seconds, a vibration 
without an impulse, during which the whip could not be released. 
Watches provided with either system of escapement, should, there¬ 
fore, in order to be used as independent seconds, beat an even 
number of vibrations per second : 14,400 or 21,600 per hour, 

either 4 or 6 per second. 


Wheel - Work. 


111 


If the watch beats 14,400 vibrations, the escape wheel ad¬ 
vances two teeth at each second ; the star of the pinion should then 
have ^ teeth ; but, since we cannot have a half tooth, we will give 
15 teeth to this piece, w r hich will amount to the same thing. 

If the watch beats 21,600 vibrations, the escape wheel advances 
3 teeth per second ; the star should have teeth, that is, 5 or a 
multiple of 5. 

One can give to the ordinary train of the watch the numbers 
of teeth generally employed. Concerning the numbers of teeth in 
the second train, we remark that, since the center wheel carries on 
its prolonged axis the second hand, this wheel should make 1 turn 
while the whip makes 60 ; one should, therefore, have 


and employing pinions with 8 and 6 leaves, 

60X8X6 = A B \ 
or, reducing into prime factors, 

2 6 X 3 2 X 5 = A B . 

Grouping these factors, one can have for example, 

A = 2 2 X 3 X 5 = 60 
B — 2 4 X 3 =48. 


The other wheels have no other condition to fulfill, except that 
the second train should run the same number of hours as the 
ordinary train, generally 32. 

The barrel, which gives motion to the train, has then also stop 
works with 4 teeth, and should make one turn in 8 hours ; that is 
to say, while the wheel carrying the second hand makes 8 X 60 or 
480 turns. One will have, then, here 


480 — 


CDE 
c d e 


Choosing pinions of 10, 8 and 8 leaves, one has 

480 X 10X8X8 = CDE , 

or 

2 12 X 3 X 5 2 = CD E , 

which gives the three groups of factors : 

C = 2 4 X 5 —80 

D = 2 6 =64 

E = 2 2 x 3 x 5 = 60. 




112 


Lessons in Horology. 


165 . If the watch has a double set of dials , that is to say, if the 
dial is subdivided into two small dials, the hour and minute hands 
of which can indicate two different times, the pinion gearing in the 
barrel of the independent second train carries a minute hand on the 
extension of its axis, as does that of the center wheel in the trains 
generally used. A set of dial wheels is added to each train, and 
one thus possesses the means of making the watch indicate simulta¬ 
neously the time of two different countries. In this case the wheel 
which has D teeth should make one turn per hour while the pinion, 
which has e leaves, carrying the second hand, makes 60. The pre¬ 
ceding figures fulfill precisely this condition, since one has, correctly, 

D E 64 X 6° 

60 = ~rr = Txr- 

166 . The arrangement of watches called fifths or quarter secojids 
is similar to that of the independent seconds ; but one could only 
construct such with an escapement whose wheel advances a half 
tooth at each vibration, since the star that is adjusted on the last 
pinion of the second train should become free at each vibration of 
the balance. One could not, therefore, employ in either of these 
systems, detent or duplex escapements. Fifths of seconds watches 
should beat 18,000 oscillations and the star of the escape wheel 
should present a tooth at each vibration ; this star should, therefore, 
have twice the number of teeth that the escape wheel has. Since 
this last generally has 15, the star should have 30. In place of the 
whip, another star, with 5 teeth, is adjusted on the axis of the last 
pinion of the second train. 

Quarter seconds watches should beat 14,400 vibrations ; the 
star of the escape pinion should have the same number of teeth as 
the wheel, and the star on the last pinion of the second train should 
have 4 teeth. 

The numbers of teeth in the other wheels are the same as for 
the independent seconds. 

167 . Let us remark that these systems are out of date to-day and 
that they are replaced by the chronographs. These mechanisms are 
simpler and consequently cost less ; they are based on entirely dif¬ 
ferent principles, having no connection with the kind of problems 
of which we treat now. 

168 . Required , to find the number of turns which one should 
give to the winding stem , on setting a watch , to make the minute 
hand move once round the dial. 




Wheel - Work. 


1 x 3 

Solution : Given the following numbers of teeth for the wheel 
in action. 

Cannon pinion, 12 leaves Minute wheel, 30 teeth 

Main setting wheel, 27 teeth Small setting wheel, 18 “ 

Sliding pinion, 16 “ (Fig. 36). 

Since it is desired to know the number of turns which the 
winding stem makes while the cannon pinion makes one, this 
cannon pinion must, therefore, be regarded as the driving wheel. 
The minute wheel, which is driven by the cannon pinion, drives in 



its turn the main setting wheel ; it is, therefore, a pinion with rela¬ 
tion to the cannon pinion considered as a wheel and a wheel with 
relation to the setting wheel considered as a pinion. The same 
thing takes place for the large and small setting wheels, which also 
drive and are driven. One should, therefore, have 

_ 12 X 3° X 27 X _ 12 _ 3_ 

71 30 X 27 X 18 x 16 16 4 ‘ 

The winding stem must, therefore, be made to execute f of a 
turn, in order that the minute hand may make 1 turn. 

One sees that the numbers of teeth in the intermediate wheels 
between the cannon pinion and the sliding pinion do not influence at 
all the result, and that the movement takes place as if the sliding 
pinion geared directly into the cannon pinion. We have, moreover, 
already established this fact when dealing with problem 159. 

169 . Let us now seek the number of turns that one should give to 
the winding stem to wind up a watch which has min a day (, 24 : hours'). 

Solution : This question deals with the calculation of the num¬ 
ber of turns which the winding pinion should make while the ratchet 
fastened on the barrel arbor makes 3. 

Admit the following numbers of teeth : 

Ratchet wheel 44 teeth Crown wheel 42 teeth 

Crown wheel, lower side, 38 “ Winding pinion 18 “ 























Lessoyis in Horology. 


1 14 


One will have, in this case (Fig. 37), 
to the preceding example 


n == 3 


44 X 38 

42 X 18 


and in an analogous manner 
6 —' turns. 

63 


It is generally desired to have this number as large as possible, for 
the reason that the effort which must be made to wind up the main¬ 
spring, being a determined mechanical work, the force which must be 



Fig. 37 


exerted to wind the watch, wall be diminished by increasing the dis¬ 
tance traversed. One sees that the number n becomes greater when we 
increase the number of teeth in the ratchet wheel and what are called 
the “crown” teeth in the crown wheel, or when we diminish the 
other teeth in the crown wheel and those of the winding pinion. 

169 a. Calculation of the Train in a Watch of the Roskopf Type. 


C 



Watches of this kind 
have a simplified train, 
inasmuch as their bar¬ 
rel gears directly into 
the third wheel. The 
movement of the hands 
is produced by the 
gearing of a wheel A 
(Fig. 37#) concentric 
with the barrel and a 
cannon pinion a placed 
on a tenon fastened at 
the center of the move¬ 
ment. The wheel A , 
moreover, carries a 
pinion b , gearing in the 
hour wheel B. The 
wheels carrying the hour 














Wheel - 


115 


and minute hands are, therefore, driven directly by the barrel. 
Let us further remark that the wheel A and its pinion should be 
adjusted to turn easily on the barrel, in order that the hands can 
be set to the hours. 

169 b. We first propose to calculate the numbers of oscillations 
of the balance in such a watch, the numbers of teeth being known. 
Suppose 


Number of 

teeth in the 

barrel. 

C 

— 

128 

< < 

< < 

i i i ( 

third wheel 

D 


84 

< < 

< < 

i i < < 

fourth “ ... 

E 

, 

60 

i ( 

i < 

n a 

escape “ ... 

F 

— 

15 




1 

r c 

— 

8 

< i 

< < 

leaves ‘ ‘ 

three pinions . . . ■< 

d 

= 

7 




( 

' e 

' — 

6 

< i 

< < 

teeth ‘ ‘ 

minute wheel . 

A 


7 2 

i ( 

< < 

< < < < 

hour “ ... 

B 

— 

66 

< < 

< < 

leaves ‘ ‘ 

cannon pinion 

a 

— 

18 

< ( 

< ( 

i i a 

minute wheel pinion . 

b 

= 

22 


The cannon pinion should make one rotation during an hour. 
We will obtain the time of one rotation of the barrel by the quotient 

A 72 
V = 78 = 4 ' 


The barrel takes four hours to execute one turn on its axis. 
The number of oscillations accomplished by the balance during one 
turn of the barrel, that is, during four hours, will be expressed by 
the formula 

C D E 'iF 


and during one hour 


N = 


CD E F 
2 c d e 


We will have, consequently, 


N = 


128 X 84 X 60 X 15 

2 X 8 x 7 X 6 


14400 oscillations. 


The train of the dial wheels will give, properly, 

72 X 66 

—-—- zrz 12 

18 X 22 

169 C. Suppose now we wish to calculate the numbers of teeth 
in the train of a Roskopf style of watch, knowing that the balance 
should make 16,200 oscillations per hour. 






116 


Lessons in Horology. 


Let us admit, as in the preceding case, that the barrel makes 
one turn in four hours. We will have 

, C D E 2 F 

16200 = -- j -. 

4 c a e 

Choosing pinions of 8, 7 and 6 leaves, one will have 
16200 X 2 X 8 X 7 X 6 = C D E F y 
and on reducing the first member into prime factors, 

2 8 X 3 5 X 5 2 X 7 = CD E F , 
with which we could form the following groups : 

C = 2 3 X 3 X 5 = 120 teeth 

D = 2 2 X 3 X 7 = 8 4 “ 

E — 2 2 X 3 X 5 = 60 

F — 2 X 3 2 =18 


i < 


< < 


The train of the dial wheel should give 

A B 


a b 


— 12. 


Choosing a — 22 and b = 18, one has 


A B = 
A B = 

from whence, for example, 

A = 2 3 X 
B = 2 X 


12 


X 

X 


22 


X 

X 


18 

II, 


3 * 

3 


X 11 


=- 72 teeth 
= 66 


i i 




CHAPTER IV. 


Gearings. 

170. Definition. The theory of gearings nas for its object the 
study of the transmission of the mechanical work from one wheel 
to another. 

171. Let us suppose, at first, that we have only one wheel 
gearing in a pinion and that in place of the complicated force of 
the spring we have a weight P (Fig. 38) acting through the medium 
of a thin and flexible cord on a 
cylinder whose radius is equal to the 
unit and which is fastened concen¬ 
trically to the axis of the wheel. 

Let us, at the same time, admit 
that the resisting force be represented 
by a weight Q suspended in the same 
manner as P from a cylinder adjusted 
on the axis of the pinion and with a 
radius equal to the unit. In further 
imagining this system animated with 
a uniform movement, the gearing will 
be perfect if, at no matter what instant 
of the movement, the work of the 
force P is equal and in the contrary 
direction to the work of the force Q> the relation of the forces 
P and Q being properly established. 

Since these forces are in the same direction as the path tra¬ 
versed by their point of application, the mechanical work effected, 
is measured by the product of the intensity of these forces by the 
distance traversed (37). 

If the relation of the forces P and Q is correctly chosen, their 
degree may be arbitrary, and, consequently, they can be supposed 
as very small or even as nothing. Therein is the basis of the 
important theory explained in kinetics. 

172. One can also exclude the movement and devote oneself 
more especially to the transmission of the force. 

We will examine the gearings from this double point of view. 



*ig. 38 


117 








118 


Lessons in Horology. 


173. Practical Examination of a Gearing. Let us place a wheel 
and a pinion in a depthing tool , in such a manner that the two movers 
may be sufficiently free, but without play between the points of the 
instrument. Regulate the distance between the two movers until the 
movement of the wheel produces that of the pinion. Impart then 
a rapid movement to the wheel : we will establish a good gearing if 
the movers conserve this motion long enough, and without any other 
noise than a certain hissing sound easily recognized. The move¬ 
ment imparted should, moreover, diminish gradually and not 
abruptly. Let us remark that, in order that this experiment may 
succeed properly, the pinion should be furnished with a wheel, per¬ 
forming the office of a “ fly, ’ ’ so that the movement may con¬ 
tinue long enough. One can also examine a gearing from this 
point of view by placing the movers in the watch and proceeding in 
the same manner. We have thus decided whether or not the 
gearing transmits the movement properly ; let us now see if it 
transmits the force correctly. 

Let us use, as in the previous case, the depthing tool, and place 
in the same manner the movers between the arms of the instrument. 
Let us then create a resisting force acting on the pinion, and, for 
this purpose, let us press tightly together the points between which 
the pinion is placed. The gearing will be found established in 
proper conditions if, after imparting a movement to the wheel, one 
feels no jerks in the transmission and has only the resistance of 
friction to overcome. 

It is necessary also to assure oneself of the “play” existing 
between the teeth of the wheel and the leaves of the pinion and of 
the proper space between the points of the teeth and the bottom of 
the pinion’s leaves. 

When the gearing is placed in the watch movement one can 
create a resisting force by pressing the end of a wooden peg against 
the end of one of the pivots of the pinion ; on causing the wheel to 
turn with the aid of another peg, one could assure oneself, as in the 
preceding case, of the qualities of the gearing considered. 

174. Let us observe that when a gearing transmits the move¬ 
ment properly, it transmits equally well the force, and when one of 
these conditions is fulfilled the other is, also. It is, however, good, 
for a careful examination, to use the two methods, for certain defects 
make themselves felt more readily by one than by the other of the 
two modes. 


Gearings . 


ng 

175. One will find that for the preceding experiments to indicate 
a good gearing, they must fulfill the three following conditions : 

i st. That the distance between the centers of rotation of the 
wheel and pinion must be exact. 

2d. That the shape of the teeth and of the leaves must conform 
to theoretical profiles. 

3d. That the total radii of the wheel and of the pinion corres¬ 
pond to the mathematical calculation. 

We would study separately each of these three conditions, 
which summarize all the mechanical theory of gearings. 

First.—Distance of the Centers. 

176. Primitive Radii. Let there be two wheels without teeth 
O and O' (Fig. 39), one driving the other by simple adhesion and 
without slipping. 



When the wheel O has turned a certain angle a while driving 
the wheel O', the point of contact a , has arrived at b , for example, 
the same point of the wheel O has then reached b' in such a way that 
arc ab = arc ab\ since the movement is effected without slipping. 

We can note, 


. . angle a O b 
1 angle a O' b' 



For two wheels having a reciprocal movement, this relation is 
precisely that of the angular velocities (34) : constant when these 
wheels or cylinders have a circular base. Moreover, if the wheel O 









120 


Lessons in Horology. 


has accomplished a number of rotations c, the wheel O' has made a 
number o' and one would have the new relation, 



ITT- Although the transmission of mechanical work by simple 
contact may not be employed in horology, at least in a direct 
manner, one finds, however, numerous applications in the work of 
the practical man. In these cases the wheels are not ordinarily in 
contact ; a certain space separates them, and to produce the move¬ 
ment of driving one by the aid of the other, we wrap around them 
both either a cord, or, perhaps, a leather strap called “ the band.” 

Thus, for example, the cord of a foot-wheel or hand-wheel in the 
watchmaker’s lathe transmits the movement, it may be, to a counter¬ 
shaft, or directly to a pulley mounted on the lathe ; the bow string 
transmits, likewise, the mechanical work produced by the hand, 
which gives motion to it, to the pulley around w r hich this cord 
is wrapped. 

178 . Let us examine, in the first place, the case of two pulleys 
connected by a cord or band (Fig. 40). Let us first establish the 



fact that, if the two sides are not crossed, the two wheels turn in 
the same direction ; if they are crossed (Fig. 41), the wheels turn 
in contrary directions. 

The angle a corresponding to 1 turn of the first pulley is 
equal to 2 ; for n turns it is 2 nr n. 

The same for the second pulley : the angle a' is equal to 2 « 
for 1 turn and to 2 nr n' for n' turns. 

One can then write 


2 nr n 

2 ir n' 



r 






Gearings . 


121 


When r and r* are known, one has for v! 


n' — n 


r 


and if, as is generally the case, n is equal to i, one has simply 



The number of turns executed by the second pulley while the 
first makes i is then equal to the relation between the radii of the 
two wheels. 



179. Applications. An arbor makes 100 tjirns to the minute ; 
it is furnished with a pulley whose diameter is equal to o.yo m. A 
band tra?ismits its movement to a pulley of 0.40 m. diameter placed 
on a second arbor . One desires to know the number of turns made 
by the second pulley. 

We will have from the preceding relation 

100 X 0.70 

7/ - 1 j 

0.40 

since 

n = 100; 2 r — 0.70 and 2 r' = 0.40, 

then, performing the calculations, 

n f — 175 turns. 

A pulley of 0.80 m. diameter executes go turns to the minute , 
what should be the diameter of the pulley driven , knowing that it 
should execute 160 turns during the same time f 

The formula 

r* _ 71 

r n' 

can be just as well written 

2 7'' n 


2 r 







122 


Lessons in Horology . 


and from thence one establishes 


2 r' = 2 r 


n 


in figures 


n 


/ » 


_ o.8o X 9° 
160 


and performing the calculations 

2 r' = 0.45 m. 

180 . It often happens that a tool, such as a lathe, a counter¬ 
sink, or a drill, should run at different speeds, in order to satisfy 
the necessities of the work. One installs then on the driving arbor 
a multiple pulley, tapered pulley or speed cone. On the driven 
arbor is likewise found a similar pulley, but always in the contrary 
manner. It is only necessary then for one cord to be placed on the 
different pairs of pulleys which correspond. The sum of the radii 
of two corresponding pulleys should then be constant. 

181 . Let us now suppose the case of a foot-lathe. The cord 
of the large wheel is wrapped around the groove of a counter-shaft 
pulley and transmits the movement to this counter-shaft. Another 
cord is wrapped around another groove of the same counter-shaft, 
but of a different radius, and transmits the movement of the arbor 
to the pulley fastened on the lathe. What is the relation between 
the number of turns of the first wheel and that of the last ? 

Let us designate in a general manner 

the number of turns of the large wheel by . 

“ “ “ counter-shaft by . 

“ “ “ “ pulley of the lathe by 

radius of the large wheel by 


71 

71 

71 

R 


tr 


4 < 


<» 


< 4 


< < 


4 4 


< < 


< 4 


4 4 


4 c 


small groove of the counter-shaft by R r 
large 
small 
large 


< < 


< < 


< < 


< < 


i« 


< < 


pulley by 
<( a 


r 

r 1 

r" 


We then have (Fig. 42) 


and, in like manner, 


n' = n 


n ff — 7 i ' 


or, replacing 71’ by its value, 


R 

R 

r 

r* 


n" = 7 t 


— V — 

R f A r /t 


Since, in this first case, the wheel drives the small pulley of 
the counter-shaft, and the large pulley of the counter-shaft drives 






Gearings. 


123 


the small pulley of the wheel, one obtains the greatest number of 
turns made by the arbor of the lathe. It is moved, then, with the 
greatest speed. 

If, on the contrary, we 
guide the cord of the large 
wheel in the large groove of 
the counter-shaft and the sec¬ 
ond cord, wrapped in the 
small groove of the counter¬ 
shaft, into the large groove of 
the pulley of the lathe (Fig. 

43), we shall obtain a lesser 
speed. 

Let us remark that, since 
it is a mechanical work which 
should be transmitted, accord¬ 
ing as the speed of the last 
pulley diminished, the force 
increases, and reciprocally. 

Thus, when one wishes to 
turn a piece of soft metal, 

such as brass, one arranges the cords in the manner to obtain a 
great speed, on condition, always, that the object to be turned is of 
small dimensions. On the other hand, if one has a hard piece of 
metal to turn, such as tempered steel, or an object oflarge diameter, 
it is proper to arrange the cords in such a manner as to obtain 
less speed. 

In the second case (Fig. 43) one has, in an analogous manner 
to the first r 



Fig. 42 


n" = n 


X 


r "' 


182. Numerical Application. Let 


n 

r 


1. 

50 mm. 


R = 400 mm. 
r’ = 20 mm. 


R' 30 mm. 
r" 40 mm. 


For the case of greatest speed, one will have 


n" = 


R X r 400 X 50 


= 33X turns. 


then has 


R ' X r ' 30 X 20 
After arranging the cords so as to obtain a slight speed, one 

R X r 400 X 20 


n 


// 


r X r " 50 X 40 


= 4 turns. 








124 


Lessons i?i Horology. 


183 . The transmission of force by the means of wheels, or 
rolling cylinders driving each other by simple contact, can scarcely 
ever be employed in practice, because the adhesion, called ‘ ‘ force 
ol friction,” is very slight; the limit being passed, slipping is 
produced. 

To obviate this inconvenience, one inserts in the wheel projec¬ 
tions, which are the teeth, gearing in the openings contrived in the 

pinion. One then forms 
what has been called the 
leaves of the pinion. With 
this arrangement the move¬ 
ment of the two toothed 
wheels should be made in 
an identical manner to that 
of the cylinders first con¬ 
sidered. 

It, therefore, follows 
that in a gearing one can 
always imagine two circum¬ 
ferences driving each other 
by simple contact, and in 
the same conditions of 
movement. These circum¬ 
ferences bear the name of 
primitive circumferences. 

184 . One calls the pitch 
of the gearing the length 
of the arc measured on the primitive circumference of one of the 
wheels, extending from a point of one tooth to the similar point of 
the tooth which follows. The pitch of the gearing should then 
comprehend the space occupied by a whole and a blank of a tooth. 

The pitch of the gearing of the wheel should be equal to that 
of the pinion which it drives. Let us designate this pitch by the 
letter p and call, moreover, the number of teeth in the wheel n , and 
the number of leaves in the pinion n’. 

The length of the primitive circumference of the wheel, 2 ^ r, 
should then be equal to p X n, since the pitch ought to be con¬ 
tained n times in this circumference. 

For the same reason the length of the primitive circumference 
of the pinion, 2 * /, should be equal to p n'. 





Gearings. 


I2 5 


In order to obtain a relation between the primitive radii and 
the numbers of teeth, let us divide the equation 


by 

we will obtain 
or, after simplifying 


2 ir r = p n 

2 ir r' = p n' 

2 ir r p n 

2 ir r' p n' ’ 


( 3 ) 




The primitive radii are then proportionate to the numbers of teeth. 

185. Calculation Of the Primitive Radii. In an exterior gearing, 
such as that which we have considered (Fig. 38), the distance 
between the centers of the two movers is equal to the sum of their 
primitive radii ; that is to say, one should have 

(4) D = r+ r\ 

D representing this distance. 

Let us take up again the proportion (3) 




> 


in which the radii r and r’ are unknown quantities and the number 
of teeth n and n’ known quantities. 

Without changing the value of an equation, one can add to 
each of its members the same term, or an equivalent term. We 
can then write 

r r f n n' 

r' r f n' n' ’ 

since the two terms -L- and dL- are both equal to 1. 

The common denominator permits us to write 

r T r ' _ n T n ' 

r' n f 


and because of (4) one will also have 

D 71 + n' 


rv 


11' 


from whence we deduce 

( 5 ) r' = D -t- 

KOJ n -f- 7 i f 

In an analogous manner we would find 












126 


Lessons in Horology. 


186. Numerical Application. A barrel of 80 teeth should gear 
in a pinion with io leaves, what should be the primitive radii of 
the two movers, knowing that the distance between their centers 
is 11.565 mm.? 

Replacing in formulas (5) and (6) the letters by their values 
above given, one will have 


r' = 11.565 X 


10 


and 


80 -f IO 


11.565 X 10 = 11565 

90 9 ’ 


r — 11.565 X 


80 _ 11.565 X 80 

80 + 10 90 


11.565 x 8 
9 


These two calculations give 


r 1 = 1.285 mm. 

r = 10.28 “ 


As a verification, one should have 

D — r + r' = 10.28 + 1.285 — 11.565. 

187 . To obtain the primitive radii, one can also simply regard 
the distance D as divided into as many parts as there are teeth in 
the wheel and the pinion together ; therefore, into n -j- n’ parts, 
and appropriate a number n of these parts as the radius of the 
wheel and a number n' for that of the pinion. The calculation is 
thus brought back to that of the preceding example. 

188 . The case of exterior gearing is the one which is most gen¬ 
erally presented in practice. 
In this system we will observe 
that the movement of the 
two mobiles takes place in 
contrary directions ; when 
the wheel is animated with 
a movement to the right, 
the pinion will possess a 
movement to the left.* 

189 . When the center 
of rotation of the pinion is 
placed in the interior of 
the wheel’s circumference 
(Fig. 44), the gearings thus 
constructed take the name 

* It is customary to call “motion to the right” that which is effected in the same direc¬ 
tion as that of the hands of a watch when looking at the dial. 












Gearings. 


127 


of interior gearings. In this case the pinion takes a movement in 
the same direction as that of the wheel. 

The distance between the centers is then equal to the difference 
between the primitive radii of the tw'O wheels. Therefore, 

(7) D = r — r f . 


If the distance between the centers and the numbers of teeth 
in the wheel and pinion are known, the value of their primitive 
radii can be calculated in an analogous manner to that which we 
have just employed to determine those of exterior gearings. We 
have the proportion (3), 

r n 

r f n ' ’ 

which can be written 


or, again, 


r 

r ' 

n 

TV 


r ' ~ 

7l' 

TV 

r — 

- r * 

71 - 

— n ' 


r' 


and, on replacing r — r* by its value D } 


from whence we find 


n 

r ' 


71 — 71' 


n 


/ » 


(8) r f = D 


IV 


71 — 71 


r 


In an analogous manner one would arrive at the conclusion 

( 9 ) r = D 


71 


71 — n' 


190. Numerical Application. Let us take as a numerical ex¬ 
ample that of a wheel with 120 teeth gearing interiorly in a pinion 
with 14 leaves, the distance between the centers being 8.75 mm. 
The application of the formulas (8) and (9) give : 


n 


r ' D -7 = 8.75 -— 

n - ^ T 20 - 


14 8.75 X 14 


and 


n — 71 

71 


120 — 14 


106 


„ _ n ~ _ o 120 _ 8.75 X 120 . 

71 — n' 120 — 14 106 


performing the calculations, one arrives at the conclusion 

r' = 1.156 mm. 
r = 9.906 “ 


The verification should always give 

D — r — r * — 9.906 — 1.156 = 8.75. 















128 


Lessons in Horology. 


191. Let us now examine a kind of gearing sometimes employed 
and which is called rack gearing. In this case the primitive circum¬ 
ference of the wheel becomes a straight line ; its radius is, conse¬ 
quently, infinite and the number of its teeth unlimited. This gear¬ 
ing can be considered either as exterior or as interior, for the dis¬ 
tance between the centers can, equally, be 

D — co -j- = co — r ' — oo. 


To determine the primitive radius of the pinion gearing in the 
rack, it is sufficient for us to know the number of its teeth and the 
pitch of the gearing. 

In Fig. 45 let a b equal the pitch of one of these gearings, 
and place 

r a b — A\ 


let us call n' the number of leaves which the pinion should have ; 
the primitive circumference will then have for its value 


which gives 


2 tr r ' = A n \ 


(io) 


A n f 

2 TT 



192. Numerical example. Let 2.8 mm. be the pitch of a rack 
gearing, the pinion must have 12 leaves, what should be its primi¬ 
tive radius ? 

The formula (10) gives 


A n ' 

2 IT 


2.8 X 12 2.8 X 6 

——;- 7 = - 7 - = 5-347 mm 

2X3-1416 3-1416 


The radius sought should then be 


r* = 5-347 mm. 















Gearings. 


129 


193. Application of the Theory of Primitive Radii to the 
Escapements. The theory of gearings finds its application not 
only in the wheel-work, but also every time that there is a question 
of the transmission of movements of rotation around two fixed axes. 
It can then be applied also in special cases, such as one encounters in 
the study of the escapements, the mechanisms of repeaters, etc. 

It sometimes happens, and especially in the last cases, that one 
knows the distance between the centers of rotation and the relation 
of the angles traversed in the same time by the movers considered, 
and that one may have to determine their primitive radii with the 
object of finding out the form of the surfaces in contact. 

The formula (1) gives us the proportion 

r a/ 

r ' a 


which indicates that the primitive radii are inversely proportionate 
to the angles traversed in any equal times. 

Furthermore, one should have, when the rotations of the two 
movers take place in opposite directions, 

D — r + r ', 


and when they take place in the same direction 


D — r — r '. 


On performing identical operations to those which we have 
indicated (185), one will arrive at the following results : 


Exterior gearing, 
and 

Interior gearing, 
and 


(II) 

r = n 

(12) 

r ' — D 

(13) 

r = D 

(14) 

r ' = D 


a + a/ 
a 

a/‘ 


a' 




194. Numerical Example. To find the primitive radii of an 
escape wheel and of the anchor, knowing, that while the wheel 
traversed an ange of io° — a, the anchor turns an angle of 9 0 = a. 
Moreover, let the distance between the centers be D — 100 mm. 

Let us remark that, the wheel being animated with a move¬ 
ment to the right, the anchor possesses a movement to the left, 






130 


Lessofis in Horology. 


when the tooth acts on the exit pallet, and a movement to the right 
when it acts on the entrance pallet. The first case is, then, that 
of an exterior gearing, while the second is similar to that of an 
interior gearing. 

The formulas (n) and (12) will give us 


and 


r = D 


r> = D 


a -f- a/ 


a -f- a/ 


100 


==■ 100 


9 9 * 

-j— — 100 —■ — 47x? 

10 + 9 19 


10 


10 19 

= 100- — 52|f 

10+9 19 


The formulas (13) and (14) will then give us 


and 



100 



100 X 9 = 900 



a 


a/ 


10 

100 -= 1000. 

10 — 9 


Second—Form of the Teeth and Leaves. 

195. General Study of the Transmission of Force in Gearings. 

In the chapter on motive forces, we compared the energy displayed 
by a motive spring to the effect produced by a weight placed at 
the extremity of a lever arm equal to the unit of distance, the 
system being in equilibrium (83). 

This fictitious weight F represents the moment of the force 
with relation to the axis around which this force exerts its action. 

By means of the gearing, this action is transmitted to the 
second axis and the problem is to find the moment F' of a force 
which, with relation to the second axis, would be in equilibrium 
with the moment F 

196. Let us suppose at first that the point of contact of the 
tooth of the wheel with the pinion leaf is found on the line of 
centers (Fig. 46), and regard the wheel O as a lever in the state 
of equilibrium. This system fulfills in effect all the conditions rela¬ 
tive to the lever ; the fulcrum is 0 } the power is F; the resistance 
is that which arises from the wheel O’ and the moment of which 
we have to find. 

This resistence is applied at the point of contact, c, of the wheel- 
tooth and the pinion-leaf ; it is directed 7 iormally to the surfaces in 
contact ; here, perpendicularly to the line of centers and consequently 
following c N It acts thus in a contrary direction to the force F 












Gearings . 


131 


The lever arm (43) of the force N is 0 c = r, its moment 
is then 


and because of the equilibrium, one should have (43) : 

(15) F — N r , 

since the lever arm of the force F is equal to the unit. 

On the other hand, the pinion is acted upon by two forces : 
one, F’, is the resisting moment to be determined ; the other, N' t 



coming from the tooth of the wheel O and acting, as also does the 
force JV, in the direction of the common normal at the point of 
contact. 

Since the pinion, as well as the wheel, is in the state of equi¬ 
librium, one should have, in an analogous manner, the equality of 

the moments : , ... .... . 

(16) /'' — N X i - 

On dividing the equations (15) and (16) member by member, 

one has „ ,, 

F _ N r 

F 7 ~ N y r r 

The normal forces N and N' are equal, since their effects 
destroy each other ; consequently, one obtains simply 

F 


r 














i3 2 


Lessons in Horology. 


from whence one finds the value sought 

(18) F ' = F ~. 
On account of the proportion (3) : 


one can then write 



( I9 ) F ' = F ~. 

n 


196 a. If, for example, the moment of F is equal to 4000 gr., 
the number of teeth in the wheel n — 80 teeth and the number of 
leaves in the pinion n' = 10 leaves, the formula (19) would become 


F ' 


10 

4000 80 ^ 500 gr ‘ 


A weight of 500 gr. suspended at the extremity of a lever 
arm 1 mm. from the center of the pinion would then make equi¬ 
librium with a weight of 4000 suspended at the same distance 
from the center of the wheel. 

Let us remark at this time that if the force has diminished 
during its transmission, and is not more, with relation to the 
pinion, than the eighth part of what it was with relation to the 
wheel, the speed of the last mover is, on the other hand, increased 
and has become eight times greater. 

197 . Supposing that the preceding calculation relates to the 
gearing of a barrel with the center pinion, let us now seek for the 
moment F" of the force that should be applied to the third wheel to 
make equilibrium with the moment of the force of the barrel spring. 

We have seen, in the preceding case, that on multiplying the 
moment F by the relation —, one obtains the moment of the force 
applied to the center wheel ; on multiplying, then, this latter value 
by the relation of the number of leaves in the third wheel 
pinion to the number of teeth in the center wheel, one will obtain 
the value sought, thus : 

n !^ 

(20) F " = F 

n n x 

198 . One could continue this reasoning for any number of 
wheels. Thus, the moment F"" that should be applied to the 
escape wheel to make equilibrium with the moment of the force 
of the spring, will be expressed by 


(21) 


F "" — F 


n ' n " n' ff n "" 

71 7i x n 2 n 3 





1 


Gear mgs. 


133 


199 . Let us choose as numerical example the very frequent case, 


F ,f " — 4000 


1 0 X 10 X iq X 7 __ 5 

80 X 80 X 75 X 70 ~ 6 Sr ‘ 


The force has become 4800 times weaker but the speed of the 
last mover is 4800 times greater. That which, in mechanics, is 
lost in force is gained in speed and reciprocally. 

200 . We have just studied the transmission of the moment of 
the force from one wheel to another, admitting that the point of 
contact of the movers is on the line of centers. 

Let us now see under what condition this point of contact can 
be found outside of that line, in such a manner that the moment of 



Fig. 47 


force transmitted preserves at each instant the same value that it 
possessed when the contact took place on the line of ce 7 iters. 

Otherwise expressed, the question is to form the teeth and the 
leaves in such a manner that the transmission of the force may be 
constant. It is necessary, therefore, that the value given by the 
formula (19) 

F ' = F — 
n 

remains the same no matter what the position of the movement. 










134 


Lessons in Horology . 


201. Let us suppose that the wheel-tooth and the pinion-leaf 
are formed in such a manner that at one instant of movement this 
contact is found at the point c (Fig. 47), situated outside of the 
line of centers. Let us find, in this position, what would be the 
value of the weight F' w r hich w r ould make equilibrium with the 
weight F f these two forces being placed at the unit of distance 
from the axes. 

The normal to the point c along which is exercised the 
reciprocal action of the tooth on the leaf and the leaf on the 
tooth, is necessarily normal both to the curve of the tooth and to 
the form of the leaf, since these two lines are tangent at this point ; 
it is directed along the straight line N N' . 

As in the preceding case, the two wheels can be compared to 
levers. The wheel O is, in effect, acted upon by two forces. The 
one, F, tending to impart to it a movement to the left ; its normal 
is F X 1, therefore, F; the other, N, directed in the opposite 
direction and arising from the pinion leaf, its lever arm being the 
perpendicular O b, its moment is 

NX O b. 

Because of the equilibrium, one will have (43) 

(22) F = NX Ob . 

The pinion is likewise acted upon by two forces : the one, F\ 
whose moment is F r ; the other, arising from the pressure that the 
tooth exerts on the leaf at the point c> following the normal direc¬ 
tion c N' ; it moment is 

N ' X O ' b '. 

Since the direction of this last force is inverse to that of F\ the 
equilibrium is produced by the equality of the moments : 

(23) F ' = N ' X O ' b '. 

Dividing equation (22) by (23), one has 

F __ N O b 
F ' ~ N ' O ' b r 

Since equilibrium exists in the system, the forces N and N\ 
w'hich have the same alignment must be equal ; in consequence, one 
has, after simplifying, 

F Ob 





Gearings. 


135 


The two triangles 

0 b a and 

O’ b ' a are similar ; 

gous sides give the proportion 



0 b 

r 


O ' b ' 

.7 > 

r 

but since (3) 

r 

n 


r ' 

n' y 

one will also have 

0 b 

n 


O ' b ' 

7l'> 

therefore, 

F 

71 


F ' ~ 

~ 7/’ 

from whence one finds the value 



F ' = 

71^ 

F -. 


n 


202 . The value of F' , identical to that which we have deter¬ 
mined in the preceding case, is then realized, and the force trans¬ 
mitted from one wheel to another will remain constant, if the 
normal common to the point of contact of the tooth and of the 
leaf passes, in no matter what position of the movement, through 
the point of tangency of the primitive circumferences. 

203 . To recapitulate, we can deduce from the preceding 
demonstrations the following rule, which is the basis for the de¬ 
termination of the forms of contact of teeth and leaves. 

In order that the transmission of force by gearings may 
remain constant , it is necessary that the acting surfaces of the 
teeth-ranges be formed by such curves that at any instant of the 
move 7 nent the normal common to the point of contact passes always 
through the same point of the line of centers , which is the pomt of 
tangency of the primitive circumferences. 

204 . It follows from this law that w’hen the contact takes place 
on the line of centers, this point is blended with the point of 
tangency of the primitive circumferences. 

205 . Let us remark that, if the normal cuts the straight 
line O O' between the points O and O ', the gearing is exterior 
and the movements of the two mobiles take place in opposite 
directions. 

If the normal cuts the straight line O O' outside of the points 
O and O', the gearing is interior, and the movement of the two 
wheels takes place in the same direction. 









136 


Lessons in Horology. 


If the normal cuts the line O O' at the point 0 \ the radius r* 

becomes nothing and one has 

F' = F — = O ; 
r 

the transmission of the movement of the force is impossible. 

If, on the contrary, the normal cuts the line O O' at the point 
0 } one has in this case r = o and consequently : 

F' — F — = 00 ; 
o 

the force F' becomes infinitely great, but the transmission of the 
movement is wholly impossible, since the primitive radius of the 
wheel is annulled. 

If, finally, the normal was parallel to the line O O' , one would 

then have F , _ ^ co F 

— 00 

This could be the case with the entrance pallet of the anchor escape¬ 
ment if the escape wheel should traverse the same angle a as the 
anchor which it drives ; one has thus (193) : 

a a a 

r = D — = r' D - = D — = 00 ; 

a — a a — a o 

the primitive radii are then infinite. 

206 . The law which we have formulated (203) shows us, even 
from the beginning, that the problem whose object is to find the 
curves of the teeth and leaves is susceptible of a great variety of 
solutions, for one may give to the teeth of one of the wheels any 
special form and find such a curve for the teeth of another wheel as 
should satisfy it, in its successive contacts with the first, according 
to the conditions given. However, the laws of the resistance of the 
materials, the wear of the rubbing surfaces, the inflexions of the 
curves, are so many causes which make us, in practice, reject the 
use of a number of these solutions. 

207 '. Let us further remark that the formula F’ = F — is 

ft 

independent of the absolute value of the primitive radii r and r * 
and depends, consequently, only on the relation of their primitive 
circumferences. 

Determination of the Forms of Contact in Gearings. 

208 . There are several methods serving to determine the bear¬ 
ing surfaces of teeth and leaves ; the basis of these different con¬ 
structions rests generally on the law which we have set forth (203). 
We will study here three of the principal of these. 








Gearings . 


137 


209. First—Graphic Method. Exterior Gearing. The funda¬ 
mental condition, that the common normal to the point of contact 
of two forms which drive each other should invariably pass through 



the point of tangency of the primitive circumferences, furnishes an 
easy graphical means to determine one of the curves, when the 
other is given. 

Let O and O' (Fig. 48) be the primitive circumferences of a 
gearing and A B the given curve of the pinion in any position. 






138 


Lessons in Horology. 


If from the point of tangency a we draw a normal to this 
curve, we will thus have the point of contact i of the leaf of the 
pinion and the tooth of the wheel corresponding to the position 
described. 

Let us remark that, in this position, the normal a i forms the 
same angle with the radius r’ of the pinion as it does with the pro¬ 
longation of the radius r of the wheel, since these two lines run 
into each other. 

Let us afterwards mark on each of the primitive circumferences 
a point, b and b ', determined in such a manner that one may have 

arc a b — arc a b'. 

Through the points b and b' draw the radii O b and O' b' , pro¬ 
longing the first sufficiently beyond the circumference of the wheel ; 
from the point b' trace the normal to the curve b' 2’, then lay off 
from the point b as summit, an angle equal to 2’ b' O’ and mark the 
point 2 making b 2 equal to b' 2'. The point 2 belongs to the 
curve sought, for if the points b and b' arrive at the position a , the 
radii O’ b' and O b will have the same alignment and the points 2 
and 2’ the same position. 

One can thus determine as many points as one wishes, and, on 
connecting them by a continuous curve, one will obtain a form such 
as D C ) possessing the ability to drive the curve A B in such a 
manner that the transmission of the movement may be uniform. 

If one conducts the curve A B in such a way that the point M,\ 
which belongs both to this curve and to the primitive circumference 
of the pinion, presents itself at the place of the point a , the point N y 
which belongs to the curve sought and to the primitive circumference 
of the wheel, should enter into contact with the point M. 

Thence it follows that one has 

arc a M — arc a JV, 

and also that when the contact takes place on the line of centers it 
is found at the point of tangency of the primitive circumferences. 

210. Interior Gearings. For an interior gearing, one deter¬ 
mines the curve of contact in the same manner as for an exterior 
gearing. 

One describes the primitive circumferences O and O’ tangent 
to the point a (Fig. 49) and the curve given A B , which we will 
suppose anew to be that of the pinion. On drawing from the point a 


Gearings. 


139 


the normal to the curve, one determines the point of contact 1 cor¬ 
responding to the position given. 

Let us indicate afterwards on the two circumferences the equal 
arcs a b and a b', a c and a c f , etc., laying off from the points 
b, c, etc., angles equal to the angles that the normals b' 2', c' 3', etc., 
form with the radii b' O ', c' O ', etc. Afterwards making b 2 = b' 2', 


0 



c 3 = c’ 3', etc., we determine the points 2, 3, etc., belonging to the 
curve sought. The only difference between this drawing and the 
preceding one lies in the fact that for the exterior gearing, one lays 
off the angles 2' b' O', 3' c’ 0 ' y etc., on the prolongation of the radii 
o b, o c ., etc., of the wheel, while for the interior gearing one lays 
these angles off from the radii themselves. 

Since we can choose arbitrarily one of the two curves and seek for 
the other, we can see that the problem allows an infinite number of 


140 


Lessons in Horology. 


solutions ; let us remark, however, that a number among them present 
inconveniences, and even impossibilities, for practical execution. 

211. Second—Method of the Envelopes. The centers of rota¬ 
tion of the two wheels are habitually fixed and the mobiles turn 
around these points. 

Let us suppose, however, that a movement of rotation may 
have been imparted to the whole system around one of the centers, 
that of the wheel, for example, and that this movement is executed 
in such a manner that its angular speed may be equal to, but in a 
contrary direction to the angular speed animating the wheel O. It 
is evident that from this method the wheel remains in a state of 
repose and that the working of this gearing will remain the same 
as if the two centers were fixed and the two wheels turned simply 
around their respective centers. 

The gearing of the fourth wheel with the escape pinion in 
timepieces called ‘ ‘ tourbillon ’ ’ offers an example of such a move¬ 
ment. The wheel is screwed on to the plate of the watch ; its 
movement is, therefore, null with relation to this plate. The escape 
pinion, pivoted in a mobile cage, turns around its center and simul¬ 
taneously with the cage, whose center of rotation is also the center 
of the fourth wheel. 

The principle of the method of the envelopes rests on this sort 
of movement. 

Let us adopt, in short, any form of leaf; on representing the 
pinion in several successive positions of its movement, around 
the wheel, we will obtain the form of the tooth, on joining by a 
tangent curve the positions that the leaf will occupy during this 
movement. 

One can then say that the tooth is the “envelope” of the different 
positions occupied successively by the leaf during the movement of 
the pinion around the center of the wheel. With this method, this 
tooth remains constantly in contact with the leaf, and the transmis¬ 
sion of the force will be effected without loss. The movement of 
the wheel being uniform, that of the pinion will consequently also 
become so. 

Let us take some examples : 

212. The transverse section of the pinion leaf of a lantern 
gearing is a circle whose center is situated on the primitive 
circumference. Suppose we wish to determine the form of the 
tooth. 


Gearings. 


I 4 I 

If (Fig. 50) the circumference passing through the points 
o, 1", 2", 3", ... is the primitive circumference of the wheel, 



and O’ A B that of the pinion, and if this last is moved without 
slipping, around the primitive circumference of the w'heel, the 






142 


Lessons in Horology. 


successive centers of the pinion will occupy in turn the points 
o', i', 2', 3', . . . On conceiving, then, the corresponding positions 
of the pinion leaf whose centers should occupy the positions 
o, 1,2, 3, . . , and on drawing the curve abed,... tangent to 

the leaf in these several po¬ 
sitions, one will obtain the 
curve of the tooth sought. 

One sees that, if the 
pinion leaf is reduced to a 
point, the form of the tooth 
would be an “ epicycloid ’ ’ 
whose generating circle 
would be the primitive circle 
of the pinion. If the leaf 
is formed by a cylindrical 
pin, the curve for the tooth 
which results from it is par¬ 
allel to this epicycloid, and is 
found removed a distance, 
equal to the radius of the pin. 

One could draw a sec¬ 
ond curve tangent exteriorly 
to the several positions that 
the pin occupies during the 
movement; the curve thus 
formed would then drive the 
pin by its concavity. 

The straight lines 1" 1, 
2" 2, . . . which connect the 
points of tangency of the 
primitive circles with the 
point describing the epicy- 
Fig. 51 cloid are normal to the curve 

at these points. 

213 . Fig. 51 shows that if the contact of the tooth with the 
leaf took place at a point such as c , before the passage of the 
line of centers, there would be produced an abutting which, 
if it did not absolutely prevent transmission of the movement, 
would, however, modify considerably the uniform transmission 
of the force. 








Gear bigs. 


143 


One knows, in fact, that the normal to the point of contact 
should pass through the point of tangency t of the primitive circum¬ 
ferences ; but one recognizes that this essential condition is not 
fulfilled in this case, since the normal cuts the line of centers at a 
point b. In place then of being (201) 

yf 

F' = F ——, 
r 


the moment of the force transmitted will no longer be expressed 
except by the value 


F' — F 


O' b 
O b' 


friction being left out. One recognizes, thus, that in lantern gear¬ 
ings, the contact of the tooth and of the leaf should commence 
very near the line of centers.* 

Note. —If the normal passed through the center O' of the 
pinion, the movement would become impossible ; if it passed on 
the other side of O', the pinion would turn in the opposite direction 


* The contact of the tooth with the leaf commences, 
in fact, a little after the passage of the center of the 
pin over the line of centers. It should commence 
exactly at the geometrical point where the epicycloid 
of the tooth turns up. . . 

Belanger, in his “ Treatise on Kinematics, indi¬ 
cates the following on this subject: 

One can determine very approximately the dis¬ 
tance between the line of centers and the point of 
inflexion, the beginning of the profile of the tooth, at 
the instant where the contact of this tooth and the 
pin commences. Let X (Fig. 52) be the point ot 
inflexion at this instant. The normal at this point, 
common to the tooth and to the piu, passes through 
the point of contact A of the primitive circles and the 
center O of the pin. The distance 0 A is at once the 
radius of the pin and the radius of curvature of the 
epicycloid WO at the point 0; let us, therefore, desig¬ 
nate it by 8. The distance A 0 is that which, in the 


formula cited in the text, 
8 = n + 


n R ' 

R' + 2 R 


is designated by n, and the distance A A, equal to 
$ — n, is very nearly the distance sought, because the 
angle of XO with the line of centers differs very little 
from a right angle (if it differs sensibly in the figure it 
is because the radius of the pin y is exaggerated to 
prevent the confusion of the lines). But the preced¬ 
ing formula gives: 



8 — n n 8 

-rT~ = R' + 2 R 2 R' + 2 R 


8 — n R’ 

from whence —=2 R' + 2 R 


Such is the relation of the 
According as one makes 


distance A A"sought to the radius 8 of the pin. 


R' 

R 


= 1, 2, 3, 4, .. 


00 




one finds 

^ — _ 1 1 3 2 1 

—g — 4 > 3» T • • ' I) 

and that the driving is effected, consequently, almost entirely after the passage of this line. 











144 


Lessons in Horology. 


to the movement indicated by the arrow, and the gearing would 
become, on that account, ‘ ‘ interior. ’ ’ 

214 . Lantern gearing can be interior and then admits of two 
arrangements, according as the interior wheel or pinion carries the 
pins and the other the teeth, or as the large wheel carries the pins 
and the pinion the teeth. 

215 . Take, again, as a second example of the application of 
this construction, the straight line A B, given as the form of the 
pinion leaf, and let it be required to determine the curve of the 
wheel tooth (Fig. 53). 

During the movement of the primitive circumference of the 
pinion around that of the wheel, the line A B will occupy succes¬ 
sively the positions A' B\ A" B'\ A ,n B'" , etc. From the points 
of tangency, 1, 2, 3, 4, etc., let us draw respectively the perpen¬ 
diculars to these lines and through the points a , b , c, d , thus 
obtained, let us make a curve pass tangent to the successive 
positions of the line A B ; we will thus obtain the form of the 
tooth. 

Let us remark that if the wheel is animated with a movement 
to the right, the position A"" B n " can become impossible for the 
transmission of the movement, for the reason that the wheel could 
then turn without driving the pinion. This shows us, moreover, 
that there exist limits beyond which the driving of the pinion by the 
wheel becomes practically impossible. 

216 . When the line A B passes through the center of the 
pinion, the curve of the tooth is an epicycloid produced by a 
point of a circle whose radius is equal to half that of the primi¬ 
tive circle of the pinion. 

217 . Note. —On comparing the two methods of determining 
the forms of contact which we have just examined, one can 
prove that the graphical method (209) is necessarily analogous to 
that of the envelopes. In short, in order to obtain the form of the 
tooth, we make one of the primitive circles with the given form 
roll around the other ; the curve sought is, therefore, in both cases, 
that which passes through the meeting point of the normals with 
the given curve, in each of its successive positions. The reason 
which has made us separate these two parts of the same whole is 
simply the greater clearness in the explanation of the subject. 

218 . Let us take, for a last sample, gearings formed by the 
evolvent of a circle. 


Gearings. 

* J 45 



Fig. 53 







14.6 


Lessons in Horology. 


The ‘ ‘ evolvent ’ ’ of a curve is another curve, such as 
C C' C" C" . . . produced by a point of a tangent to the first 
curve, whose contact changes continually, in such a manner that 

the distance of the de¬ 
scribing point from the 
point of contact may 
be constantly equal to 
the space traversed by 
the point of contact on 
the curve. Thus (Fig. 
54), B' C\ B" C ", . . . 
being positions of the 
tangent, one should have 
B'C' = B'C ; B"C" = 
on which the tangent rolls 



Fig. 54 


vt 


B " C, etc. The curve C B' B’ 
is the ‘ ‘ evolute ” of CC r C" . . . 

The point C where the evolute meets its evolvent is the origin. 

219 . For gearing formed by the evolvent of a circle , one adopts 
for the motive tooth the evolvent of any circle concentric and 
interior to the primitive circle of one of the two wheels. The 
profile of the corresponding tooth for the other wheel will be 
determined, then, in a very simple manner. 

Let the evolvent D' D' of the circle E’ E’ (Fig. 55) be given. 
In order to determine the point of contact M of this curve and of 
the form sought, draw the normal A A' from the point of tangency 
of the primitive circumferences ; by the construction this normal is 
at the same time tangent to the “evolute” circle E ' E'. But if 
from the centers O and O' we draw the perpendiculars O B and 
O' B 1 on this normal, we will obtain two similar triangles whose 
homologous sides are in the same relation. But the sides a O , 
a O' and O' B’ are constant, being the radii of invariable circles ; 
therefore, O B must be also constant. Consequently, the normal 
A A ' of the curve D D sought remains always at the same distance 
from the center O and it, therefore, envelopes a circle E E con¬ 
centric and interior to the primitive circle of the w r heel and whose 
radius is found with that of E' E’ in the same relation as those of 
the primitive circles themselves. The form of the tooth is, therefore , 
another evolvent of a circle. 

The point of contact being found at any instant on the line 
A A\ this line is the geometrical place. 








Lrearings , 


H7 



Fig. 55 














148 


Lessons in Horology. 


Thus, in gearings of this kind the place of the points of con¬ 
tact is the common tangent to both circles of construction. 

The common normal retains, therefore, a fixed position in 
space during the movement of the two wheels around their respec¬ 
tive centers. 

This right line can make any angle with the line of centers ; 
it is the general custom, however, to place them at 75 0 from each 
other. 

The especial advantages of this system of gearings are, first, 
that the two wheels being similar and the teeth not showing any 
change of curvature at the passage of the line of centers, any one 
tooth will drive the other before as well as after the line of centers. 


Moreover, the construction of a wheel not depending in any'' 
on that which it should drive, all wheels evolvents of circles 
can gear together ; the relation of the velocities which they have 
is only to be considered. This is a valuable property which 
allows a single motive wheel to drive at once several others, or 
to make several wheels gear together successively, as is the 
case in the screw-cutting lathe. Another advantage to be con¬ 
sidered is that the distance of the centers can vary between cer¬ 
tain limits without the regularity of the gearing suffering in 
consequence. 

The gearing of evolvents can be interior ; the form of the 
teeth, in place of being convex, is then concave. This fact is an 
inconvenience which makes this combination little used. One can, 
in these cases, diminish the concavity by multiplying sufficiently the 
number of teeth. 

220. Third—Roller Method. The principle of this method 
differs from the preceding, but is just as general. 

Let us imagine, first, any polygon, AD C D E F G (Fig. 56), 
compelled to roll without sliding the length of a line X Y. At a 

certain moment of the movement 
one of the angles, A, for example, 
is found in contact with the line 
X Y. During the rolling around 
this point all the points of the 
polygon, and with them all those 
which, interiorly or exteriorly, could 
be unalterably connected with them, 
describe arcs of circles around the 







Gearings. 


149 


center A. As, for example, the point H, exterior to the polygon 
but unalterably connected with it, will describe an arc a b during 
the instant of the rolling considered. 

The radii of these diverse arcs of circles will be their normals 
and will necessarily pass through the point A. 

Let us remark that the length of each arc described depends 
on that of its radius and on the number of sides that compose the 
polygon. If we suppose this geometrical figure formed with a great 
number of sides, the lengths of 
the arcs described while it turns 
around one of its sides, will 
diminish. At the limit, that is to 
say, when the number of sides 
becomes infinite, the polygon is 
confounded with a continuous 
curved line, and each of the 
points which compose it will de¬ 
scribe, nevertheless, as the poly¬ 
gon rolls around an instantane¬ 
ous point of contact, an infinitely 
short arc of a circle. But, how¬ 
ever small it may be, this arc pos¬ 
sesses, nevertheless, two extreme 
radii, drawn infinitely near to 
each other and passing through 
the instantaneous center of rota¬ 
tion. Since they are drawn infi¬ 
nitely near to each other, either 
of these radii of curvature is, consequently, normal to the point 
considered of the total curve described by this point during the 
continual rolling of the generatrix along the line of the directrix. 

This established, let there be, moreover (Fig. 57), any curve, 
a b c, which we cause to roll on the exterior of the primitive cir¬ 
cumference of a wheel and at the interior of that of the pinion. 
If, to be more clear, we suppose that a point H taken outside of 
this curve mav be connected with it in an invariable manner, the 
movement of this point will be similar to that of all the points 
composing the given curve. 

During a certain period of the curve’s movement at the 
exterior of the primitive circle of the wheel 0 , this point H will 





Lessons in Horology. 


150 


describe a trajectory d Hg ; then, when the movement takes place 
at the interior of the primitive circle of the pinion O ', its trajectory 
will be the line f H e. These two curves can be adopted as the 
profile of conjugate teeth. In fact, we imagine that the curve a b c 
follows the movement of the two primitive circumferences in such a 
manner that these three curves remain constantly tangent at a. 

The trajectories meet at H> since 



this point describes them both ; 
moreover, they are tangent there, 
since the normal for each is ob¬ 
tained on joining the describing 
point H to the point of contact a 
of the moving curve a b c with 
both of the primitive circumfer¬ 
ences established. 

Consequently, the common 
normal of the teeth, at their point 
of contact, passes through the 
point of tangency of the primitive 
circumferences, and the verifica¬ 
tion of this fact suffices, we know, 
in order to have the curves ob¬ 
tained, adopted as forms of teeth. 

Let us examine from this 
point of view the following case : 

221. Flank Gearings. In or¬ 
der to obtain a profile very much 
used in the practice of horology, 
one chooses as the generating 
form the circumference whose 
diameter is the radius of one of the primitive circles and one 
takes the describing point on its circumference (Fig. 58). 

In the movement of the generating circle around the primi¬ 
tive circle of the wheel, the point A describes an arc of an 
‘ ‘ epicycloid ” AD. 

In its movement in the interior of the primitive circle of the 
pinion, this same point A describes a straight line O' A , which is a 
radius of the circle O’. This plane surface O' A is called a “flank." 

Let us remark that the epicycloid which forms the profile of the 
tooth in flank gearings is not the same as that which we have 







Gearmgs. 


151 


determined for the lantern gearings (212). In the first case it is 
produced by a point of a circle with a radius less than one-half 
that of the primitive circle of the pinion, and in the second this 
curve is produced by a point of the primitive circumference itself. 

222. We are now going to prove that in the rolling of the 
interior of the circle with twice the radius , the moving point tra¬ 
verses a diameter. 

If one represents in effect any position whatever, O" } of the 
moving circle during its movement in the interior of the primitive 
circumference of the pinion, the angle inscribed, A' O' M, has for its 
measure the half of the relation of the arc A’ M comprised between 
its sides to the radius A' 0 \ that is to say, 

One can, on the other hand, measure it as an angle to the 
center O' by the relation of the arc comprised A' A to the radius 
A' O' ; therefore, 


But, if the expression of the theorem is true, that is to say, if 
the point A of the generating circle is carried to M along the 
straight line A O', the two angles A' O' M and A' O' A should be 
equal and superpose ; we would, therefore, have the equality of 
the terms : 


A' M 
A' O' 


A' A 
A' O'' 


The arc A' M is equal, in fact, to the arc A' A , since the 
rolling of the generating circle is effected without slipping ; the two 
relations are, therefore, equal and the point M is found, in conse¬ 
quence, on the radius A O'. 

Since it relates to any instant whatever of the movement, this 
point, therefore, does not leave the diameter O A, which is, then, 
properly the trajectory sought. 

223 . If one imagines the flank in any position whatever, as, 
for instance, O' D (Fig. 59), its point of contact M will be obtained 
by erecting to it the perpendicular A M. The angle A M O' 
being a right angle, the point M will be found on the circum¬ 
ference which has A O’ as diameter ; consequently, in flank gear¬ 
ings, the location of the points of contact is the generating circum¬ 
ference itself. 

224 . An analogous reasoning to that which we have de¬ 
veloped for a preceding case (213), shows that in the simple 
flank gearings the driving can only take place on one side of the 

line of centers. 




152 


Lessons m Horology. 


225 . In order that the contact of two similar teeth may com¬ 
mence before the line of centers and end on the other side of that line, 
it suffices if each tooth has a mixed profile formed with a flank interior 

to its primitive circle and 
with an epicycloidal part 
exterior, generated by a 
circle with a diameter 
equal to the radius of the 
primitive circle of the 
other wheel. Thus, for 
example (Fig. 60), the 
circle O" furnishes in its 
successive rollings a flank 
O f A for the wheel 0 ’ and 
a curve A D for the 
wheel O. The circle 0 l " 
furnishes in an analogous 
manner a flank O A for 
the wheel 0 and a curve 
A D' for the wheel O. 

This combination is 
called “reciprocal” flank 
gearing. One can, there¬ 
fore, say that in recipro¬ 
cal flank gearmgs the 

driving takes place on 
both sides of the line of 
centers. 

Let us add that the 
form of reciprocal flank 
jQ. gearings cannot be em- 

F ig. 59 ployed for interior 

gearings. 

226 . Two wheels with plane interior flanks and epicycloidal 
curves exterior to the primitive circles should, according to the 
generation of their profiles, be made especially for each other, since 
a wheel cannot gear regularly in several others of different diame¬ 
ters. This inconvenience is avoided for a series of wheels that one 
wishes to make gear with the same wheel, by replacing in the wheels 
of the series the straight flanks by curves, one chooses for generat- 




Gearings. 


153 


ing circle of these interior curves and of the corresponding exterior 
curve of the particular wheel, a constant circle whose diameter dif¬ 
fers the least possible from the radii of the wheels of the series. 

One encounters in horology an example of this case in the 
gearings of the dial wheels and the setting wheels. The cannon 
pinion drives the minute wheel, in which also gears the main setting 
wheel ; this drives, in its turn, the small setting wheel (168). 



An inverse movement is produced when the hands of the watch 
are set to the hour, and it is then the small setting-wheel which 
drives the other wheels. 

One can, in this case, take the circle O" half of the primitive 
circle of the cannon pinion, as generating form of the exterior 
epicycloids of the wheels and afterwards make this same generating 
circle roll in the interior of each of the primitive circumferences 
considered, in order to obtain the interior form of the teeth, this 
form is then a “ hypocycloid ” (Fig. 61). 




J 54 


Lessons in Horology. 


In practice, one substitutes very often straight lines for these 
hypocycloids, and thus one obtains a general outline recalling that 
of the flank gearings, although incorrect from the point of view of 
its construction. 



226 a. Determination of the Profile of a Tooth Corresponding 

to a Proiiie Chosen Arbitrarily (according to Reuleaux). Suppose 
given a b c d e f g . . . i j k the profile chosen, A and B the primi¬ 
tive circumferences of the two wheels whose respective centers are 
O and O' (Fig. 61 a). 

One draws the normals a 6 , b 5 , c 4 , . . . 2 h , 3 z, 4 y, 5 fc, to the 
profile given. Through the points a, b, c y d . . . i, j , k, one passes 
arcs of circles described from O' as center. From S’ as center with 








Gearings. 


155 


the lengths a 6 , b 5 , c 4 . . . 4 j, 5 k (normals) one describes arcs of 
circles which will determine the intersections VI, V, IV, III, II, I, 
I\, Hi » This series of points, connected, form the 

line of the gearing (place of the points of contact). This done, 
from the point O as center, one describes arcs of circles passing 
through the points I, II, III, . ... f, If, III X , . . . The lengths 
of arcs Si, 1.2, 2.3, 3.4, 4.5, taken on A, will be retaken on B 
and will determine the lengths of the corresponding arcs 6 /j, 



1 1 2 X , 21 3,, j, 41, 4xS 1 (instantaneous centers of rotation). If 
from these last one lays off the lengths of the normals a x 6 X = a b, 
b x 3 x — b3, c 1 4 X — c 4, etc., and if these are connected by a con¬ 
tinuous curve, one will have the profile sought, a 1 ,b l ,c 1 . . .i l ,j l k 1 . 

226 b. Gearings by the Evolvent of a Circle. In extension of 
that which we have said about evolvent of circle gearings (219), 
one can further establish, in a very simple manner, the kind of 
generation of the forms of contact by employing the method of the 
rollers (220). 

In effect, the primitive circumferences of such a gearing being 
known and the generatrix being the straight line A A f (Fig. 55) 










Lessons in Horology. 


156 

inclined on the line of centers and passing through the point of 
tangency of the primitive circles, the rolling of this line around a 
tangent circle E E , interior to the primitive circumference of the 
wheel, will cause to be described by a point M of this line an 
evolvent D D. The rolling of the same line around a circle 
tangent, but interior, to the primitive circumference of the pinion, 
will cause to be described by the same point M a second evolvent 
D' D ', which is the form of the conjugated tooth. 

It is clear that the rotation of a line tangent to a circle is 
effected in the same manner when, according to the condi¬ 
tion established, it must be accomplished on the exterior of the 
wheel’s primitive circle and on the interior of the pinion’s primi¬ 
tive circle. 

Remark : Let us further state the fact that an evolvent of a 
circle is nothing more than an epicycloid described by a point' of a 
generating circle whose radius is infinite. 

Teeth-Range. 

227 . Up to the present, before approaching the details relative 
to the distribution of the teeth on their wheels, we have been 
occupied solely with the determination of the curves, or profiles 
of contact, by which these teeth mutually drive each other, without 
determining the points where they terminate. The time has now 
come to pay attentian to these questions. 

From the geometrical point of view, a single tooth could, 
strictly speaking, suffice for the transmission of the movement ; 
but, in practice, there would result complications and physical 
impossibilities, independently of the obstacles also very serious, 
arising from friction. One furnishes, therefore, the wheels with 
several teeth, and it is, for this reason, necessary to make them 
all identical. 

Each tooth has two profiles. Strictly speaking, the posterior 
face could be left any shape ; such are, for example, the teeth 
called “wolf,” in some gearings for stem winders. It happens, 
however, often enough, in mechanics, that sometimes one wheel 
drives another, and sometimes it is driven by the other ; therefore, 
the movement takes place in both directions. It is best, for this 
reason, to construct the two faces alike. The tooth is then 
“symmetrical” with relation to a radius of the primitive circle, 
which is, in some degree, its ‘ ‘ bisectrix. ’ ’ 


157 


The teeth being identical and their number a whole number, 
they, therefore, divide the primitive circumference into a certain 
number of equal parts between them, which is called the “ pitch ” 
of the gearing (184). This pitch is subdivided into three parts, 
the full , the blank and the play. The full is the space measured 
on the primitive circle and occupied by the material of the wheel ; 
the blank is the surplus, which should remain clear to allow the 
introduction of the conjugate tooth ; the play is an accessory blank, 
which does not seem at first to be necessary from the geometrical 
view-point, but which, in reality, is indispensable. 

Numerous causes render the play necessary for the action of 
gearings : for instance, imperfections in the divisions of the wheels 
by the machine, the shake necessary to the pivots in their holes, the 
expansion of the bodies of which the mobiles are formed and the 
inevitable introduction of foreign bodies in the wheel teeth, are all 
so many reasons for this necessity. 

228 . It can be said that the play is the relation of the arc not 
occupied by the sum of the breadths of the tooth and of the leaf, 
to the pitch of the gearing. 

If we represent by p the pitch of the gearing, by a the length 
of the arc occupied by the tooth on the primitive circumference 
and by b the length occupied on this same circumference by the 
corresponding tooth of the other wheel, the play j will be expressed 
by the formula 

p — a — b 
J ~ P 

If, for example, we had for a given case 

p — 6 mm., a = 2.8, and b — 2.8, 

one would obtain the play of the gearing 

6 — 2.8 — 2.8 _ 1 
J = - 6 - ” T?> 

229 . To determine the quantity of play necessary for a gear¬ 
ing, one has to examine two conditions : First the solidity of the 
wheel teeth and then the space to be reserved for the free passage 
of small foreign bodies, such as dust or the particles which, inevi¬ 
tably, are detached from the bearing surfaces on account of wear. 
One or the other of these conditions can have the predominance, 
according to the nature of the gearing. 





I 5 » 


Lessons in Horology. 



Thus, in the gearing of stem-winding works, of the setting 
gear, the rack of a repeating watch, etc., the conditions inherent 
to the solidity of the wheel teeth should evidently predominate. In 
the gearing of the fourth wheel with the escape pinion, there must 
necessarily be reserved space for foreign bodies. 

The same two conditions should also guide us in the choice of 
the form to give to the part of the wheel teeth which forms what is 
called the depth of the teeth. Thus, when one desires a solid set 
of teeth one chooses in preference the rounded depth, as Fig. 62. 

If, on the contrary, one wishes 
to construct a set of teeth leav¬ 
ing place for foreign bodies, 
one will adopt a form such as 
is indicated in Fig. 63 ; in this 
last case, one can also use the 
lantern gearing. 

230 . In the gearings of 
stem-winding mechanism, of 
change wheels for setting, etc., one divides the play equally between 
the two wheels ; in the gearings of the train of the watch one 
deducts it from the pinion 
leaf alone, for the reason that 
this last mobile is made of 
tempered steel and offers, 
consequently, more resistance 
than the brass of which the 
wheel is made. The solidity 
of the leaf is, furthermore, 
also increased by its greater 
transverse length. 

Another reason which makes us deduct the play from the leaf, 
is that the wheels generally drive the pinions ; consequently, it is 
the profile of the tooth which drives the flank of the leaf after the 
line of centers. The curve of the tooth, therefore, must be of 
sufficient length to be able to drive the flank far enough to prevent, 
as much as possible, the tooth following entering into contact before 
the line of centers. This, however, is not always possible for the 
pinions of low numbers. 

231 . For the mobiles of the train, the general rule adopted is 
to give in the wheel half of the pitch to the tooth and the other 



Fig. Go 






Gearings. 


159 


half to the blank. The pinions of 6, 7, 8, 9 and 10 leaves would 
then have one-third of the pitch appropriated to the leaf and 
two-thirds to the blank. In the pinions of 12 leaves and above, 
one would give two-fifths of the pitch for the breadth of the leaf 
and three-fifths for the blank. 

Thus the gearings with pinions of 10 leaves and below have 
a play of 




= etc., 


and the pinions of 12 leaves and above 



12 — ^12 — §.I2 
12 



232. For the gearings of the change wheels, one can admit 
of play. 

233. For those of the stem-winding mechanisms, one can be 
content with ^ of play. 


Third—Total Diameters. 

234. Before entering into the details relating to the determina¬ 
tion of the total diameters of the mobiles in a gearing, we will 
commence by the geometrical study of the curves employed in 
horological gearings. The principal among these we know to be 
the epicycloid. As a preface to this question, let us establish, first, 
the theory of the cycloid. 

Cycloid. 

235. Definition. The cycloid is a curve described by a point 
of the circumference of a circle which rolls without slipping along 
a straight line. 

This curve is employed in the rack gearings (191), which 
establish a connection between a uniform transfer and a uniform 
rotation around an axis perpendicular to the transfer. This is then 
the particular case of gearings around two parallel axes in which 
one of the primitive circles, having its radius infinite, becomes a 
straight line. 

236. Drawing Of the Cycloid. Let it be desired to describe 
by points the cycloid generated by the point A of a circle with 
diameter D (Fig. 64). 

One draws a straight line A A' equal to the base D of the 
cycloid. One describes the circle O with the diameter D tangent at 





i6o 


Lessons in Horology. 


the point A to the line A A '. One divides the generating circum¬ 
ference and the base into the same number of equal parts, 12, for 
example, which are numbered in the manner indicated by the figure. 
From the point of the center O one draws a straight line parallel to 
the base ; this line will contain the successive places transversed by 
the center of the generating circle during its rolling. Let us indicate 
on this parallel the positions of the center O, corresponding to the 
positions of the circle when it is in contact with the base at the 
division points, and, from each of these centers, let us describe 
circumferences with diameters equal to that of the generating 
circle. 

Let us remark that when the generating circle has arrived at 
the center / 0 , the point i x of its circumference is lowered to 1 , and 
the point A , whose movement describes the cycloid, should be ele¬ 
vated to a height equal to the distance which the point i 1 is lowered. 

Thus, in this new position, the point A should be found on 
the circumference whose center is at 1 0 and also on a line parallel 
to the base passing through i l . 

In the same manner, we could determine the successive 
positions occupied by the point A while the center of the gene¬ 



rating circle is found at 2 0 , j 0 , ^ 0 , etc., and on connecting all 
the points thus obtained by a continuous line, one obtains the 
cycloid ABA' sought. 

237 . Drawing of the Cycloid of a Continuous Movement. One 

understands that the circle O is a circular plate on the circumference 
of which a point or pencil A is fixed (Fig. 64). If one causes the 
plate to turn without slipping along a straight rule whose edge 
coincides with A A', the point or pencil A will describe the cycloid 
of a continuous movement. 



















Gearings . 


ibi 


238. Normal and Tangent to the Cycloid. Let M be any 

point whatever on the cycloid A M B (Fig. 65), through which 
it is desired to draw a normal, then a tangent. Having traced 
the base A A' and its parallel E E' containing the places occupied 
successively by the 
rolling, we will find 
the center O of the 
generating circle 
corresponding t o 
the point M of the 
eycloid, by tracing 
from the point M 
with an opening of 
the compass equal 
to the radius of the 
generating circle, an 
arc of a circle pass¬ 
ing through the line 
E E '. The point 
of intersection O will be the center of the generating circle. 

Dropping from the point O a perpendicular on the base A A r , 
the point P will be the momentary center of rotation of the gene¬ 
rating circle. Its movement is composed of a movement of trans¬ 
lation parallel to the base and of a movement of rotation around 
its center O. The point P being thus the center of this combined 
movement, the point M will describe an arc of a circle infinitely 
small around this momentary center ; the straight line M P being 
the radius of this arc will consequently be the normal to the point 
M\ sought. The tangent being perpendicular to the normal, should 
pass through the point T of the generating circle ; one knows, in 
fact, that every angle inscribed in a semi-circumference is a right 
angle. 

239. Evolute and Radius of Curvature of the Cycloid. The 

evolute A\ A of the semi-cycloid A B is a semi-cycloid equal to 
its evolvent (218). 

Let A A' be the base (Fig. 66), B f the axis of the cycloid 
generated by the point M of the circumference T M P. Let us 
describe a circumference on the diameter P P' = T P; through 
the point P' draw E F parallel to A A' ; then draw the lines MM\ 
M T and M' P f . On account of the equality of the angles MP T 


centers of the generating circle during the 









i 62 


Lessons in Horology . 


and P' P M *, the two right-angled triangles P M T and P M' P' 
are also equal and one has 


from whence 


M f P = P M 
M M' — 2PM 


The straight line M M' is the radius of curvature of the point 
M; that is to say, the radius of the circumference which has two 


T B 



consecutive elements infinitely small, common with the curve at 
this point. 

Thus the radius of curvature at any point whatever M of the 
cycloid is twice the portion M P of the normal comprised between 
the curve and the base. 

Designating the angle MOP by a , by 8 the radius of curva¬ 
ture and by r the radius of the generating circle, one has 

8 = 4 r. sin \ a 

It is thus easy to see that the foot A r 1 of the evolute corres¬ 
ponds with the summit B of the evolvent, while the summit of the 
evolute blends with the origin of the evolvent. 

240. Length Of the Cycloid. The length of the portion A M' 
of the cycloid A M ' A\ is equal to the length of the radius of 
curvature of the point M of the cycloid A M B ; it is equal to 












Gearings. 163 

M M' t since this is the length of the line unwound from the cycloid 
portion A M'. 

One, therefore, has 

A M' = M M' = 2 M P= 4 r. sin \ a. 

In order to obtain the length /' of the cycloid portion A\ M’ 
we have evidently the difference : 

l' — 4 r — 4 r sin \ a = 4 r (1 — sin \ a). 

Epicycloid. 

241. Definition. The epicycloid is a curve described by a 
point of the circumference of a circle rolling without slipping on 
the circumference of another circle. 

The generating circle can either roll on the exterior or on the 
interior of the director circle ; in the latter case the interior epicycloid 
is called hypocycloid. 

We have seen that this curve is employed for the form of teeth 
in the gearing of two wheels turning around tw^o parallel axes. 

242. Drawing Of the Epicycloid. This drawing is analogous 
to that of the cycloid. Let us describe first from the center C the 
director circumference on which the generating circle O should roll. 
Mark on the circumference C a length A A' equal to the length of 
the circumference of the generating circle O. The latter being 
tangent to the point A } divide its circumference and the base A A' 
into an equal number of parts, 12, for example. 

From the center C describe afterward a circumference with a 
radius CO) on this circumference will be the places occupied, 
successively, by the center of the generating circle ; draw then the 
radii C A, C lt C 2) . . . etc., prolonged to the circumference pass¬ 
ing through the center of the generating circle. Describe then 
from the points i Q} 2 0 , j> 0 , . . . etc., as centers, circumferences 
with radii equal to the radius of the generating circle. 

Note now that w r hen the center of the generating circle has 
arrived at i 0 , the point 1 x of its circumference is lowered to 1 ; this 
point has, therefore, approached C the same distance that the 
point A has been removed from it. On describing, therefore, from 
the center C a circumference passing through the division i x of the 
generating circle, we obtain the point A x by the intersection of 
this last circumference with that of the generating circle from the 
point 1 q. 


164 


Lessons in Horology 


In the same manner we could determine as many points as we 
wished, and on connecting them by a continuous line we would 
obtain the epicycloid sought, as it is represented in Fig. 67. 

243. Drawing of the Epicycloid of a Continuous Movement. 



Fig. 67 

C and O being circular plates and A a pencil point fixed in the 
circumference O, one understands that on making the plate O roll 
without slipping on the plate C, the pencil will trace the epicycloid 
A B A' of a continuous movement. 

244. To Draw a Normal, then a Tangent to the Epicycloid. 

Identical considerations to those which have enabled us to draw a 






Gearmgs. 


5 


normal and a tangent to the cycloid furnish us the means of drawing 
these same lines to the epicycloid. If it is desired to draw a normal 
to the point M of the epicycloid A MB (Fig. 68), we commence 
by seeking the center of the generating circle belonging to the 
point M of the epicy¬ 
cloid, by laying off 
from this point Af, the 
radius of the gene¬ 
rating circle, on the 
circumference around 
which it is moved dur¬ 
ing the rolling of its 
center. From the 
point O l we figure 
the position of the 
generating circle and 
we find the point of 
tangency P, of the 
two circumferences. 

During an infi¬ 
nitely small period of 
movement all the 
points of the gene¬ 
rating circle, as also 
those which are un¬ 
changeably connected 
with them, describe 
infinitely small arcs of 
circles. Since the nor¬ 
mal of an arc of a 

circle unites with the radius, we would have, therefore, the radius 
M P as normal to the point M of the epicycloid A M B. 

On drawing afterward from the point Af a perpendicular to the 
normal, we would obtain the tangent M T. This tangent should 
pass through the point T of the generating circle, since every angle 
inscribed in a semi-circumference is a right angle. 

245. Evolute and Radius of Curvature of the Epicycloid. We 
have seen that the evolute of a cycloid is a cycloid equal to its 
evolvent and placed in such a manner that the summit of the 
evolute coincides with the origin of the evolvent and the summit of the 






Lesso?is in Horology. 


166 


evolvent is found on the same perpendicular to the base as the origin 
of the evolute. Moreover, the two bases are parallel to each other 
and separated from each other the diameter of the generating circle. 

As regards the epicycloid, we will see, moreover, that its 
evolute is a similar curve, but not equal. The summit of the 



\ ; . / 
' » : # 

* c • ' 


\\ •/ 
vv 

c 


Fig. (if) 


evolute and the origin of the evolvent at the point A (Fig. 69) 
still coincide ; the summit B of the evolvent and the origin A x " are 
found on the same radius B C , but their generating circles are of 
different diameters. 

The two demi-epicycloids are contained in the same angled C B 
which we will designate by 0. If we call R the radius of the base of 
the evolvent and r that of its generating circle, R' the radius of the 











Gear mgs. 


167 


base of the evolute and P that of its generating circle, we see that 
the length of the base of the evolvent is 


and that of the evolute 


r 0 — ir r 

R' 0 = ir r'. 

On dividing one of these equations by the other, one obtains 


or 


R 0 

R' 0 

R 


ir r 
ir?’ 


R' P' 

On the other hand, one should still have 

R — R' = 2 r', 

R — R' 

r' — 

on substituting 


from whence one draws 


P . = 

R 


2 

2 r 


and 


R' R — R' 
(1) R' : 


R 1 _ 

R -f 2 r 


Let, for example, R — 60 mm. and r — 20 mm., we would 
have, in this case, 

60 2 _ 3600 


R' = 


60 -f- 2 X 20 


100 


36 mm. 


and 


. 60 — 36 

r' — -— = 12 mm. 


246 . The point M’ (Fig. 69) is the center of curvature of the 
point M of the epicycloid A M B ; it is situated on the evolute 

A x " M’ A. 

We have, in effect, 

arc M P = arc A P, 

since the generating circle has rolled without slipping on the cir¬ 
cumference A A'. But, 

arc M P — r a and arc A P = R X angle A C P 

Let us call the angle A C P, 7 and place 

ra = Ry, 

one will then have 


r 

~R 


JV_ 

a 











i68 


Lesso?is in Horology. 


When the generating circle with radius P has rolled without 
slipping on the base B' A 1 " } this length of arc B' Af is equal 
to ir r'. One has also 

it X = arc P M' + arc M' P', 

then 

arc P' M' = arc P' A 
arc PM— arc B' P', 

and as 

arc P M' = P X angle P O' M', 
arc B' P' — R' y t 

one will then have 

r' X angle P O' M' = R' y, 

from whence 

r' _ y 

R 7 ~ angle P~O r M r ' 

But as we have 

P _ r 

IB' ~ ~R' 

we will also have 

y _ _ y _ 

a angle P O' M' ’ 


from whence a = angle P O' M'. 

The point M’ thus determined belongs, therefore, properly to 
the evolute. 

Since the angle M T P = M'P' P = ^ a , and since the angles 
at M and at M' are right angles, the straight lines MP and PM' 
will have the same alignment. 

247 ". The straight line M M’ representing the line developed 
is the radius of curvature of the point M of the evolvent and the 
length of the arc A M' developed. 

We have, in effect, 

MM' = M P + P M '; 


or 

M P — 

therefore, 


2 r. sin \ a and P M' = 2 P. 
M M' = 2 (f + r') sin \ a. 


sin \ a ; 


Designating the radius of curvature by 8 and replacing P by 

R — R' R / _ R \ 

2 ~ 2 V R + 2 r)' 


we will have 

(2) 8 = 2 r ( 


1 + 


R 


R -f 2 r 


R + 2 r 
^ sin ^ a 


4 r 


R + r 
R + 2 r 


sin 


For a numerical example, let r = 20 mm., P = 60 mm., 
a — 6o° ; we will obtain successively, 










R 

R + 2 l 


Gear bigs. 


169 


then 

+ 


60 


2r ( 


60 -f- 40 


= 0.6 and 1 -j- 


R 


1 + 


R 


R + 2 r 


R -f- 2 r 
.j = 40 X 1.6 = 64. 


1.6. 


Log. 64 = 1.80618 

“ sin i a = 9.69897 

Log. 6 = 1.50515, from whence 5 = 32 mm. 


For y 2 a = 90°, the radius of curvature 8 = A 1 ", B = 64 mm. 

248. We know that in flank gearings, the radius of the gene¬ 
rating circle of the epicycloid is equal to half the primitive radius 
of the pinion which we will designate by r' ; one has, therefore, in 
this case (2), 2 r __ r / 


The radius R takes, then, the notation r, primitive radius of the 
wheel. The angle ^ a is the angle formed by the flank of the 
pinion leaf and the line of centers. Under these conditions, the 
formula (2) becomes 

8 = r' ( 1 + TqT^T-) sin \ a = r' l 1 + i + sin \ a ; 

or, again, n and n' being the numbers of teeth (184), 



249. The formulas (2) and (3) show that at the origin, the 
radius of curvature is nothing. This fact indicates that at this 
point the curve is united to the primitive radius of the w r heel 
without forming an abrupt angle. 

The radius of curvature increases, afterward, proportionately 
to the sine of the angle formed by the flank of the leaf and the line 
of centers ; it becomes greatest when the angle ]/ 2 a is equal to 
90° ; it diminishes then to become again zero for yi a = o, that 
is to say, at the point of the curve’s inflection. 

250. Length of the Epicycloid. The length of arc of a curve 
is equal to the length of the line developed. Thus (Fig. 69) 

arc A M f = MM f 2 r ^ 1 -f- s i n \ a - 

One has most frequently occasion to determine the length of 
an epicycloidal arc calculated from its origin ; it is, therefore, 
necessary to determine its length from the point A x n . For this 
purpose it is equally proper to take in place of the angle P P' M f 















Lessons in Horology. 


170 


its complementary P' P M' , which we will designate by \ 0. For 
the point M" the angle \ p becomes thus equal to P" A" M" which 
obliges us to change the sine into cosine. 

To obtain the length A x " M" of the epicycloid A 1 " M " A> 
let us remark that this length is equal to A A x " — A M". 

Therefore, 

/= 2 K i + JTF^)- 2r ( I+ xTir) cosi? ’ 

/ = 2, '( : + ('- tosi O ; 

r is here the radius of the generating circle of the epicycloid A MB 
and R the radius of its base ; so we have (245) 

r — ( R' -J- 2 r' ) and R = R / 2 r '. 

On substituting these two values in the above equation, we 
will obtain the length of the epicycloidal arc, calculated from the 
point of origin, thus : 

(4) l = 4 P ( 1 4- 1 — cos \ p). 


251. First Application. A wheel of 80 teeth can drive the 
leaf of a pinion with 10 leaves, after the line of centers, an angle 
\ p = 34 0 45' 48" (257) ; what is the length of the epicycloidal 
arc of the tooth, starting from its origin when the primitive radius 
of the wheel is 10 mm.? 

Solution : The radius X of the generating circle of the epicy¬ 
cloid is X 10. We have, therefore, 

yf 

I + -FJ7 = 1 -f tV = 1.0625 

and 


4 r* — 4 X t§ = 2.5 


from whence one obtains 



2.5 X 1.0625 = 2.65625. 


The natural expression of cos \ p is 


it follows 
Then 


Cos \ p = 0.821514, 

1 — cos \ p — 0.178486. 

log : 4 ^ ( I + -^7) = 0.42427 
+ log : (1 — cos \ p ) =0.25160 — 1 

log : l — 0.67587 — 1, 


from whence 


l = 0.4741 mm. 










Gearings. 


171 


Remark. —The height of the ogive is equal, in this case, to 
o. 4 22 85 mm. We indicate further on, the means of calculating 
this latter value (258). 

252. Second Application. A wheel of 60 teeth can drive the 
leaf of pinion with 6 leaves, an angle ^ p = 4 2 ° 15' 47" after the 
line of centers ; what is the length of the epicycloidal arc of the 
tooth, the primitive radius of the wheel being 5 mm.? 

Solution : We have here r* — 0.25. Therefore, 


1 + 


and 


R' 


— 1.05 


4 r' = 1 


COS \ P 


0.74006 
cos \ p =0.25994. 


Consequently, 

log : 4 *’ ( 1 + -^7) — o.o 2 H9 
+ log : (1 — cos \ p ) = 0.41487 — 1 

log : / = 0.43606 — 1 
l — o. 2 7 2 93 mm. 


Remark. —The height of the ogive is, in this case, equal to 
o. 2 3 2 5 mm. 

253. Third Application. Similar problem for the gearing of 
a wheel of 70 teeth in a pinion with 7 leaves, the angle \ p being 
39° 55' 1 5" and the primitive radius of the wheel 5 mm. 

Solution : We have 


log : 4 r 1 1 + = 0.0:2119. cos } p = 0.76693 

log : (1 — cos 1 p ) = 0.367 4 9 — 1 1 cos -g- p = 0^3307 

log \ l — o. 38868 — 1 

and l = o. 2 447 2 mm. 

Remark. — The height of the ogive is, in this case, 
o. 2 ii 65 mm. 

The calculation is, therefore, the same for all the flank gearings, 
it is useless to follow further examples of the application. 


Relation of the Radius Vector to the Angle formed by the Variable 
Radius Vector and the Initial Radius Vector. 

254 . The radius vector C M (Fig. 70) which we will designate 
by 5, forms with the initial radius vector C A — K, radius of the 
base, an angle 0 ; one can conceive that there should exist 








172 


Lesso?is in Horology. 


a relation between the radius S and the angle 0. This relation is 
complicated, but it has a great importance in the calculations 

relative to the de¬ 
termination of the 
; total radius of the 
; wheels. 

1 

If the angles 
M O' A' = a and 
M C O’ — p are 
known, we would 
have the propor¬ 
tion 

8 



Fig. 70 


and in the second 
from whence 


r 


sin a 
sin p ’ 


from whence 

c sin a 

5 — r — —Q-. 
sin P 

Let us now 
seek for a relation 
between the angles 
0 and a, and for 
this purpose pro¬ 
ject the point M on 
the straight line 
C O'; we will thus 
form two right- 
angled triangles 
ME O’ and ME C. 
In the first we have 

M E = r. sin o, 


M E = (R + r — r. cos a) tang, p, 
r. sin a = (E -f- r — r. cos a) tang. p. 
On dividing by r, it becomes : 

sin a = —f- 1 — cos tang. p. 

But 

A A' = M A\ 

therefore, 

r ft = JP (0 + p): 






Gearings. 


173 


from whence 
then 

( 5 ) sin a 



255 . Remark.— One can also project the point O’ (Fig. 70) 
on the prolongation of CM, and one thus forms the two right- 
angled triangles O’ H M and O' H C'. In the first case, we have 


O' H — r. sin (a -f- 3), 

and in the second 

O' H = {R + r) sin 0; 

from whence 

r. sin (a -j- (3) = (7? -f r ) sin p. 

On replacing a by its value 

a = A(0 + p) 

and on dividing by r. it becomes 

sin [A. (0 + p) + p] = 0L + i ) sin p, 
which one can also write 

(6) sin [A- 9 + (A + 1 ) p ] = (~ + 1 ) sin p. 


256 . The calculation of the equations (5) and (6) is compli¬ 
cated ; one sees, in fact, that one can only proceed by successive 
approximations. We give below an example of this kind of 
calculation. 

25 ?. Numerical Application. To find the value of the angle a 
(Fig. 70) corresponding to the position of the point M in the epicy¬ 
cloid of the tooth of a wheel with 60 teeth gearing in a pinion with 
6 leaves. 

We will suppose that the point M considered belongs to the 
point of the tooth ; it is, therefore, the extreme point of the curve 
of this tooth. 

The application of the formula (5) gives us, first, 


R 

r 

Moreover, 


+ 


60 r x 

= t + i = 2 .;-^ = -^ = o.o 5 . 


360° 

4 X 60 


i° 30', 


since half of a tooth should take up a quarter of the pitch of the 
gearing. 




i74 


Lessons in Horology. 


Let us suppose, first, the angle a = 8o° ; we would then have 

P = a — 0 = 4 0 — i° 3</ = 2° 30'. 

K 

The calculation gives 

R R 

- --J- I = 21 log : — -J- I - COS a = 1.31861 

— cos 8o° = 0.17365 -f log : tang : p = 8.64009 

-b 1 —cos a = 20.82635 9.95870 

r 

The logarithm of the second member of the equation (5) is, 
therefore, 9.95870 ; in order that equality may exist between the 
first and second members, it would be necessary for the above 
logarithm to be equal to that of sine a = sine 8o°. We have 

log : second member = 9.95870 
log : sin : 8o° = 9-99335 

difference = 0.03435 


The equality of the two members of the equation (5) is not 
verified ; the logarithm of sin a is too great by 0.03465 ; it is con¬ 
sequently necessary that the value of a be greater than 8o°. Let 
us try to take a = 86°. We will have 

P = —7^- a — 0 = 2° 48'. 

A 

R | . R 1 

— + I = 21 log : — + I — COS. a = 1.32077 

— cos: 86° = 0.06976 log: tan: p = 8.68938 

20.93024 10.01015 

log : sin a = log sine 86° 9.99894 

Difference, .01121 


This time, the logarithm of the first member is smaller than 
that of the second, which indicates that the angle of 86° is too 
great. The angle a should, therefore, be found between 80 and 
86 degrees. 

Let us now establish the following proportion, taking note that 
for 6° of arc the difference between the natural values of the cosines 
is 0.04586 : 


0.04586 6 

O.OII2I x 

on making the calculation 


from whence x — 


6 X 1121 
' 4586 ’ 


x = i.° 4666 .= i° 28'. 

Thus, if the difference which one obtains by the calculation of 
the two members were proportionate to the difference of the angles 















Gearings. 


175 


TABLE SHOWING THE ANGLE \ a TRAVERSED BY THE PINION OF SEVERAL 
ORDINARY GEARINGS DURING THE CONTACT OF A TOOTH OF THE 
WHEEL WITH THE LEAF OF THIS PINION. 


Number of Teeth 

Angle of Driving 
after the 

Line of Centers 

Angle of Driving 
before the 

Line of Centers 

Total 

Driving Angle 

Wheel 60 . . . 

Pinion 6 . . . 

42 0 15' 17" 

17 0 44' 13" 

6o° 

Wheel 70 . . . 
Pinion 7 . . . 

39 ° 55 / 15 " 

11° 30' 27.857" 

51 0 25' 42.857" 

Wheel 60 . . . 

Pinion 8 . . . 

37° 36' 20" 

7° 23' 40" 

45 ° 

Wheel 64 . . . 
Pinion 8 . . . 

37 0 42' 30" 

7 0 17' 30" 

45 ° 

Wheel 80 . . . 

Pinion 8 . . . 

38° o' 55" 

6 ° 59 ' 3 " 

45 ° 

Wheel 75 . . . 

Pinion 10 1 . . 

34 ° 39 ' 33 " 

1° 20 7 7" 

36 ° 

Wheel 80 . . . 

Pinion 10 . . . 

34 0 45' 48" 

1° 14' 12" 

36 ° 

Wheel 90 . . . 
Pinion 12 ... 

32 0 27' 30" 


32 0 27' 30" 
in place of the 
30° necessary 

Wheel 96 . . . 
Pinion 12 . . . 

32 0 33' 14" 


32 0 33' 14" 
in place of the 
30° necessary 

Wheel 120 . . . 

Pinion 12 ... 

32 0 5 °' 


32 0 5 o' 

in place of the 
30° necessary 























































176 


Lessons in Horology. 


chosen, which is not entirely the case, the angle of 86° would be 
i° 28' too great, which gives a new value for a, say, 86° — i° 28' = 
84° 32'. Let us begin again, therefore, the verification for this 
last value. One has 

a = 84° 32' and consequently p = 2 0 43' 36" 

R R 

-(- I = 21 log : — + 1 - COS a = 1.32024 

— cos a = 0.09527 + log : tang : p = 8.67784 

20.90473 9.99808 

log : sin a = 9.99802 

Difference, 0.00006 


One sees that we have very nearly approximated the real 
value of a ; if it is desired, one can approach it still nearer by a 
new approximation and one arrives at length at a = 84° 31' 34" 
and p = 2 0 43' 34". 


Calculation of the Total Radius of the Wheel. 

258 . Knowing, now, the two angles a and p, it becomes easy 
to calculate the value of the total radius of the wheel ; that is to 
say, of the radius 8 (Fig. 70), ending at the point of the ogive of 
the tooth. (See preceding calculation : 257.) 

Making this calculation for the primitive radius R = 1, we 
would have : 1 


r — 


20 


R — 0.05 R 


for the data of the preceding calculation. 
One should then have (254) 


and 


8 — 0.05 


sin a 
sin P 


log : sin : a = 9.99801 
log : sin : p = 8.67728 


1.32073 

-f log : 0.05 = 0.69897 — 2 

log : 8 ==✓ 0.01970, from whence 8 


1.0464. 


Form of the Excess of the Pinion Leaf in a Flank Gearing. 

259 . The gearings of the wheels in a watch always turn in the 
same direction,* and it is the teeth of the wheels that drive the 
pinion leaves. 


* Except, however, the setting wheels, which are driven by the pinion when the watch 
runs. In this mechanism, the minute wheel does not drive the cannon pinion except at the 
moment when this train is turned by the hand in setting the watch to the time. 












Gearings. 


177 


The equation (5) can be written under the following form, 
remarking, however, that the primitive radii are proportionate to 
the number of teeth and that, in this formula, the radius r of the 
generating circle is equal to half of the primitive radius of the pinion. 


sin 0 = + 1 “ cos - a ) tan s- (*ir a - 0 )• 

When in a given gearing one finds through the calculation of 
the above equation, the angle 

360° 


\ a 


n 


/ » 


the tooth of the wheel may drive the pinion leaf a sufficient quan¬ 
tity after the line of center, so that the contact of the following 
tooth may commence on this line. Such is, for example, the case 
of a wheel with 96 teeth gearing in a pinion of 12 leaves. 

We have, in fact, in this case, 

a = 65° 6' and \ a == 32 0 33'. 


The tooth can, therefore, drive the leaf an angle of 32 0 33'. 
On the other hand, we see that the angle which separates two con¬ 
secutive flanks is 


Consequently, one can prove that the tooth drives the leaf 2 0 33' 
farther than is absolutely necessary. 

On the contrary, for the gearing of a 6o-tooth wheel in a 6-leaf 
pinion, one has <*• = 84° 31' 34", from whence \ a = 42 0 15' 47". 
Moreover, 



Since the leaf should be driven by the tooth an angle of 6o° 
and it is driven in reality only an angle of 42 0 15' 47" after the line 
of centers, the contact must necessarily commence 17 0 44' 13" 
before this line, since 42 0 15' 47" -f- 17 0 44' 13” = 6o°. 

The gearing should be arranged, in this case, in such a 
manner that the tooth enters into contact with the leaf before the 
line of centers, and we have seen that, for this purpose, the leaf of 
the pinion must be terminated with an epicycloidal form susceptible 
of being driven by the flank of the tooth up to the moment when 
the contact is made on the line of centers ; starting from this point, 
it is the curve of the tooth which drives the flank of the leaf (225). 







i 7 8 


Lessons in Horology. 


The epicycloid of the leaf should be described by a point of a 
generating circumference whose diameter is equal to the primitive 
radius of the wheel. 

260 . Since the angle at which the tooth should enter into con¬ 
tact with the leaf before the line of centers is never very consider¬ 
able, it is very rarely necessary that the excess of the tooth be per¬ 
fectly ogival. This shape would be, moreover, rather hurtful than 
useful, because the friction would be increased and it would also 
necesitate longer teeth for the wheel, in order to allow the free 
introduction of the pinion leaf in the space which separates two 
teeth of the wheel. 

One preserves, therefore, only that part of the epicycloid 
which is directly useful. In large mechanics, one simply removes 
the desired quantity from the points of the teeth.* 

In horology, one terminates the pinion leaf by a rounded 
form, an arc of an ellipse, for example. This shape of the excess 
should be determined in such a manner that its radius of curva¬ 
ture at the point of connection with the curve of the epicycloid, 
should be the same for the two lines (Fig. 71). It is evident 
that, for security, one makes the contact of the tooth and the leaf 
commence some degrees sooner than is necessary. Thus, in the 
preceding case, one will admit a contact commencing 20° before the 
line of centers, rather than 17 0 44' 13". 

261 . Practically, in a great number of ordinary pinions, one 
finds the excess of the leaf terminated by a half circle. We are 
really in a position to recognize that this form is defective, espe¬ 
cially for pinions of low numbers. It is easy to prove that with 
this shape, the tooth drives the leaf too far after the line of centers, 
that the point of contact does not rest on the circumference of the 
generating circle and that there can be established, before the line 
of centers, a contact of the flank of a tooth with the rounded part 
of the leaf. There results from this last act a butting often very 
pronounced. 

It is probably this butting observed by horologists, which 
frequently makes them exaggerate the importance of the friction of 
the gearings before the line of centers and which has made them 
believe that this last is much more considerable than that which 
is produced after this line. The friction observed should only be 
in reality the butting, and it suffices to suppress this to per- 


*This operation is called the chamfering of the teeth. 





Gearings. 

ceptibly diminish the difficulty 
of the driving. The calculation 
proves, in fact, that the differ¬ 
ence between what is called the 
‘ ‘ entering ’ ’ friction and the 
* ‘ leaving ’ ’ friction is not great 
enough to be so easily found. 

262 . In order to determine, 
geometrically, the form of the 
excess of a pinion leaf, it is, 
therefore, established that this 
form should be composed of an 
epicycloidal arc sufficiently long 
to be able to enter into contact 

• 

with the flank of a tooth of the 
wheel several degrees before the 
point of the succeeding tooth 
leaves the circumference of the 
generating circle. Two teeth can, 
therefore, drive at once two leaves 
through a small arc. 

The symmetrical epicycloids 
of the leaf are afterward termi¬ 
nated by an elliptical shape con¬ 
necting them. Suppose, then, it 
be desired to determine this curve, 
which has no other condition to 
fulfill except that its radius of 
curvature at the connecting point 
must be the same as that of the 
same point of the epicycloid. 

263 . Let us determine, first, 
the radius of curvature of the 
epicycloid at a point correspond¬ 
ing to a driving of 20° before the 
line of centers for a wheel with 
60 teeth gearing in a pinion with 
6 leaves. 

Since the generating form 
of the epicycloid of the leaf is 



» • 



179 


i8o 


Lessoyis in Horology. 


a circle with diameter equal to the primitive radius of the wheel, 
the formula (248) 

8 = r> ( 1 + r ^ sin a 


r J 


r \ r' 


^ sin \ a. 


Q 


20 


V 


should be written here 

S = r ( 1 + 

Admitting 

r = 6o, r' — 6 consequently \ a = 60 = 2°, 
we would then have 

8 = 60 ^ 1 + sin 2 0 = 65.454 X sin 2 0 

and, making the calculation 

8 = 2.28433. 

264 . Let us now seek the value of the angle w, formed by the 
radius of curvature and the straight line passing through the middle 
of the leaf (Fig.71). 

The angle a c O is a right angle and we have, therefore, 

a c O — 90°. 

The angle a O c, or \ a, is 2 0 , 

a O c = 2 0 ; 

it follows that 


/ L 


and 


c a O — 88° 

O' a c = 180 0 — 88° = 92 0 . 
We have, moreover, the angle 


a O' d, — 20 0 — 


3 6o c 


2X3X6 


= io°, 


since in this case the width of one leaf is equal to one-third of the 
pitch of the gearing (231). 

We will have at length 

w — c d s — O' da — 180° — O' a c — a O' d 
w — 180 0 — (92 0 -f- io°) 
w = 78°. 

265 . We have afterward to determine the straight line O' c y 
joining the center of the pinion to the first point of contact. 

In the triangle G c O' we know the sides 


G O' = \ r + r' 
G c = \r 

O' G c = 4 0 . 


and the angle 







Gearings. 


181 


We will have, consequently, the value of the angle a 0 f c ex¬ 
pressed by 


tang : a 0 / c= - 


\ r sin 4 0 


or 


\ r r' — \ r cos 4 C 
sin 4 0 

tang : a O' c — . r' 

+ i t- — cos 4 °. 


The calculation gives 


a O' c — 19 0 o' 46 // . 


The side O' c of the triangle O' c G will be at last given us by the 
formula 


\ r. sin 4 0 
sin a O' c' 


from whence, after making the calculation, 

O' c = 6.4236. 

266 . Let us further project the point c on <?, on the straight 
line O' s, designating by y the right line c e and let us determine 
this line. We will have 


but 

c O' d = 
therefore, 

and 


y = O' c sin c O' d , 

a O' c — a O' d = 19 0 o' 46 // — io° 
y = 6.4236 X sin 9 0 o' 46" 

y = 1.00626. 


= 9 0 o' 46"; 


26 T. We now know the radius of curvature 5 at the point c 
of the epicycloid, the inclination of this radius to the line passing 
through the middle of the leaf, therefore the angle w and, finally, 
the ordinate y corresponding to this same data. There only re¬ 
mains, now, to determine an elliptical curve capable of satisfying 
these conditions. 

268. Radius of Curvature of an Ellipse. The equation of this 
curve being 

a * + b* x* = a 3 b 2 , 

one obtains 

d y b 2 x 

d x a 2 y 

and 

d'y b 4 

dx 2 a 2 y 2 ‘ 

The general equation of the radius being 



d x 2 













i 82 


Lessons in Horology. 


we will obtain successively 

/ b 4 -t 3 \* (a 4 y -f 6 4 -^ 3 )* 

8 = L^_ g4 y ' J . = (^y) 1 A4 

° £ 4 -^ 



ta‘S )» 

= (« 2 ^) 3 = a e y , 

therefore 


(a 4 y )* 

a 6 JJ/ 3 « 

= -— — = a : 


a 2 y 

a jj / 3 

consequently 


8 = 

- ( a* y -j- b 4 )* 

a 4 £ 4 


We have the length of the normal MN (Fig. 72) expressed by 



and 


y V 1 + (t £> 2 = * V 1 + % 


x‘ 


4 y 2 


JT vW’ + *• ** ^7 = ^/iV 

M N — l/ g * T* + - 1 * 

a? 

MN , = w + *• *• .. ) » . 


One can, therefore, place 


8 = 


a 


a* M N 3 


we have, on the other hand, 


M N — 


y 


sin w 


6 = 


a’y* 


b 4 sin 3 w * 


therefore 


























Gearings. 


183 


from whence 


(1) a’ = 


5 b 4 sin 3 2 v 

y 


In order to determine the value of b, let us remark that 


P N = 


y 


tang zv ’ 

P N being the sub-normal, the general equation of which is 

b 2 x 


c d y b 2 x 

Sn = y ~J- X = — y 


from whence 
and 


y 


tang, zv 


a y 

b 2 x 
a 2 


a 


a > 


a y 

— x — -=— 


b 2 tang zv' 

Substituting the values of x and of a 2 in the general equation 

a 2 y 2 + b 2 x 2 = a 2 b 2 , 

which we can write under the form 


We will have 


a 2 y * 


8 sin 3 w 8 sin w cos 2 


zv 


a 2 b 4 tang 2 zv y tang 2 zv 
from whence 

2 ,, / 8 sin zv. cos 2 w 

y 


y 


and 


(2) 


= b' ( 1 — 


) 


r 


y — 8 sin w. cos 8 zv 
The equation (1) gives, moreover, 


(3) a = b‘- 


V 


8. sin 3 zv 


These two last values are those of the semi-axes of the ellipse, 
which fulfills the conditions sought. 

269. Numerical Application to the Preceding Example. We 
will have 

y = 1.00626. zv = 78°. 8 = 2.28433; 

consequently 


1.00626 


and 


.00626 — 2.28433 s i n 78° cos 2 78 c 


= 1.05834 


a 


= 1.05834 * J^g 43 3 ; in ; 78! _ i . 6224S . 
\ 1.00626 3 























184 


Lessons in Horology. 


270. Total Radius of the Pinion. The total length of the 
pinion leaf is composed of the sum of the two lengths O’ e and 

ep (Fig- 70 - 

The distance ep is equal to a — x (Fig. 72). We have 


in figures 


and 


— x = 


a* y . 

b 2 tang w ’ 


— x 


1.62245 2 X 1.00626 
1.05834“ X tang 78° 


0.50266, 


a — x = i. 62245 — 0.50266 = 1.11979. 


On the other hand, we have (Fig. 71) 


we know (265) 
and the angle (266) 
therefore 


O' e = O' c. cos c O' e , 
O' c = 6.4236 
c O' e — 9 0 o' 46 // ; 


O' e = 6.4236. cos 9 0 o' 46"= 6.3444. 
The total radius of the pinion will be 


O' e = 6.3444 
+ (a — x) — 1.1198 
R' — 7.4642. 


271 . Note. —If the excess of the above pinion leaf was formed 
by a semi-circumference, we would obtain its total radius in a much 
more simple manner. 

The primitive radius being 6, the length of the circumference 

is as follows : ..... . . . 

primitive circumference = 2 ir X 6. 

The pitch of the gearing should be equal in length to 2 *■, since 
the number of leaves is 6. The width of a leaf, reckoned on the 
circumference, is one-third of the pitch, therefore 

width of a leaf = ; 

3 

finally, the height of the excess is half of this last value, therefore 


TT 

excess = —. 

3 

We will have, consequently, the total radius, 


R' = r' + 
or 

in round numbers. 


- = 6 + 1.0472 = 7.0472 
R' — 7-05 






185 


Gearings. 


Graphical Construction of Gearings. 


272 . Let us suppose that we know the distance between the 
centers D, also the numbers of teeth of the two mobiles, and let 
these be, for example, D — 240 mm .; n = 70 teeth and n' — 7 
leaves. 

The formulas (185) 


r = D 


n 


n + n f 


and r' = D 


n' 


n 


H- n' ’ 


give us the value of the primitive radii of the two wheels 


r ----- 240 


70 


70 + 7 


and r' 


240 


70 + 7’ 


the calculations made, one obtains 


r = 218.18 .... and r' = 21.818 .... 

From the centers O and O' (Plate I), previously determined, 
one describes the two primitive circles calculated, tangent at the 
point a 

Let us suppose that we wish to construct a flank gearing, one 
can commence by determining the shape of the tooth. Describe, 
for this purpose, the generating circle O" with radius equal to half 
that of the primitive circle of the pinion, and by its rolling around 
the primitive circle of the w r heel, let us make it describe the 
epicycloid A B. 

Let us remark that for the clearness of the plan, instead of 
putting the origin of the curve on the line of centers, as would be 
the logical w r ay to do, we have carried this construction back to 
another point, which, practically, amounts to the same thing. 

In order to construct the epicycloid, one can follow the method 
we have indicated (242), or proceed more simply by carrying back 
on several successive positions of the generating circle, such as 
1, 2, 3, ... , certain lengths ij I — 1*0, 2 1 II = oi x -f ij 2 lf 
3j III = oij + ij 2, + 2 1 3,, etc. On connecting the points 
I, II, III, ... . by a continuous curve, one obtains the epicycloid 
of the tooth. 

This method, which is not exactly correct, since it substitutes 
the lengths of chords for the lengths of arcs, is, however, admissible 
for drawings in which the successive positions of the generating 
circle are relatively close together. 

We divide, afterwards, one of the primitive circumferences, 
that of the pinion, for example, into as many parts as that wheel 






Lessons in Horology. 


186 

should have teeth. We can commence this division at any point of 
the circumference, but generally it is commenced on the line of 
centers, or at the point where the leaf of the pinion should be found 
at the moment of the last contact with the tooth. 

In order to determine this last position, one can make use of 
the table (257), which gives us, for the gearings most generally 
used in horology, the angle of driving of the leaf by the tooth 
after the line of centers. Thus, for the gearings of a wheel of 70 
teeth with a pinion of 7 leaves, we see that the tooth drives the 
leaf 39 0 55' 15" after the line of centers. 

If one wishes to determine this position, graphically correct, 
a previous division of the primitive circle of the pinion gives 
the “pitch of the gearing.” One lays off this length from the 
origin of the curve A , to C; one divides the pitch into two 
equal parts, one of which should be occupied by the whole of the 
tooth, the other by the space (231). The extremity D of the 
ogive of the tooth will be determined by drawing the prolonged 
radius O D passing through the middle of the tooth. From the 
center O , one describes an arc of a circle passing through the 
point D and one thus obtains the point d, extreme position of the 
contact of the tooth with the leaf (223). One draws, afterwards, 
the straight line O' d prolonged to the primitive circumference of 
the pinion. The point i is then the point of departure for the 
definite division of the pinion. If one has proceeded with exacti¬ 
tude, the angle a O’ i should be, in this case, equal to 39 0 55' 15". 
One can thus prove that the first contact of the tooth with the leaf 
should take place before the line of centers and at an angular dis¬ 
tance from this line of at least n° 30' 27". 

The excess of the leaf should be formed with an epicycloidal 
arc, as we have already indicated (259). This epicycloid is that 
which is described by a point of a circumference with radius equal 
to half the primitive radius of the wheel. 

The curves of the teeth being thus formed and their positions 
determined, it becomes easy to construct the gearing, by remarking 
that for the gearings of watch trains, the leaf of the pinion occupies 
the third of the pitch for pinions of 10 leaves and less, and two- 
fifths for those of 12 leaves and more (231). 

One then limits the epicycloid of the leaf, according to what we 
have said, by conserving to it only a length sufficient for the driving 
to commence a little before the point where the first theoretical 


Gearings , 


187 











Lessons in Horology. 


188 

contact should be effected, normally. During a very short instant, 
two consecutive teeth are thus simultaneously in contact, and this 
fact suffices to insure the correct action of the gearing. 

The leaf is afterwards terminated by an arc of an ellipse whose 
radius of curvature at the junction of the two curves is the same as 
that of the epicycloid determined. 

One then limits the length of the flanks of the leaves and teeth 
by arcs of circles with radius sufficient to allow not only the free 
introduction of the teeth and the leaves in the corresponding spaces, 
but also reserving the place which foreign bodies would occupy, 
dust and other matters which are invariably introduced, with time, 
into the sets of teeth. 

The gearing is thus constructed and having made the drawing 
on a sufficiently enlarged scale, one could deduce from it all the rela¬ 
tive dimensions for its practical construction, as we will see later on. 

273 .* Plate II represents the same drawing to a still more 
greatly enlarged scale; the distance of the centers is 2200 mm., 
the primitive radius of the wheel 2 meters and that of the pinion 
200 mm. This design allows us to show more clearly the manner 
in which the contact of the tooth with the leaf is effected before the 
line of centers; the shape of the leaf, represented in dotted lines, 
is semi-circular ; one sees thus that in this case the normal at the 
point of contact does not pass through the point of tangency of the 
primitive circumferences, as is the case for the semi-elliptical shape, 
and consequently the force transmitted has not the value which 
we determined (195 and the following) 

cv zr n ' 

F' — F — 
n 

and that there should be produced a “butting.” 

274 . The drawing of gearings of ratchet wheels, setting wheels 
and dial wheels, etc., is executed in an analogous manner ; we will 
examine later on the several modifications admitted for such wheel 
teeth. In this construction there must also be taken into considera¬ 
tion the manner in which the “play” is distributed (232, 233). 

Practical Application of the Theory of Gearings. 

275 . In practice there are presented problems of different 
natures in which it is desired to determine the relative dimen¬ 
sions of wheels and pinions. 

* As Plate II has, for lack of space, been reduced one-half, the distance of the centers is 
1100 mm.; the primitive radius of the wheel, 1 m., and that of the pinion 100 mm. 





( 189 ) 






















190 


Lessons in Horology. 


It is evident that, at first sight, the use of a suitable instru¬ 
ment to establish these sizes becomes very important to the work¬ 
man, for the reason that it saves him all calculation. We will cite 
the one which is the most exact and at the same time the most 

simple to use. 

2T6. The Proportional 
Compass and its Use. The pro¬ 
portional compass, in its most 
rational arrangement, is formed 
of two rule plates, straight, and 
divided into equal parts ; they are 
joined together at one of their ex¬ 
tremities by means of a hinge O 
(Fig- 73)- 

These rules can be fastened in 
any position by means of a clamping 
screw V 

Their divisions should be ex¬ 
actly corresponding, equal to each 
other and numbered. The point 
zero is found at the hinge, summit 
of the angle b O b'. 

The proportional compass is 
based on the fundamental principle 
of similar triangles, in which the 
homologous sides are proportional. 

Thus, imagining the primitive 
diameters of a wheel placed at the 
division of the compass correspond¬ 
ing to its number of teeth and the primitive diameter of the pinion 
at the division corresponding to its number of leaves, one should 
have the proportion 

b b' o b 



Fig. 73 


a a' 


o a 


Since, in a gearing the number of teeth of the mobiles should 
be to each other as their radii, or their primitive diameters, one 
understands that to determine the primitive diameter of a pinion, 
knowing that of the wheel, it suffices to place the latter at the 
division corresponding to its number of teeth and, for this purpose, 
to open the two arms of the compass the proper distance. The 












Gearings. 


191 


primitive diameter of the pinion should then coincide with the 
division which corresponds to its number of leaves. The proportion 


r' 


n 

n' 


is then found to be verified. 

But, as has been shown before, we run against the difficulty of 
not being able conveniently to fit the primitive diameters of the two 
mobiles in the compass, since these diameters are only theoretical. 
The difficulty has been overcome in the following w'ay : 

277 . On dividing the primitive diameter of any wheel by the 
number of its teeth, we obtain a length which we call “ diametrical 
pitch ’ ’ of the gearing. The proportional compass always gives 
the diametrical pitch by its division 
1 when the wheel is placed so that ..*••• 

its primitive diameter corresponds in 
the instrument to the division of the 

number of its teeth. it , \ \ : 

But, if we measure the height . 

of the ogive a b (Fig. 74) and, on 
account of the one which is oppo- 
site, we double this value, if we after¬ 
ward divide this figure by the dia- % . ... 

metrical pitch, we obtain a quotient Fig. 74 

which, added to the number of teeth, 

will give the total diameter of the wheel in units of diametrical pitch. 
This diameter is then 

. 2 a b 

n + — 3 ~, 

d being the pitch considered. 

On now placing the total diameter of the wheel at the division 

. 2 a b 

n + — T~' 

its primitive radius will be by this fact placed at the division 71. 
The same for the pinion. 

2 <T 8 . Example. Let us consider a wheel with 60 teeth gearing 
in a pinion with 6 leaves, and let us represent graphically this 
wheel with a primitive radius of 54° mm * 

The diametrical pitch should be 

2 X 54o 


60 


18 mm. 






192 


Lessons in Horology. 


Let us describe the epicycloid of the tooth by making a gene¬ 
rating circle with radius equal to half the primitive radius of the 
pinion, roll around the primitive circle of the wheel ; r' being the 


radius of the pinion. 


One will have 

w = 6 X 54 Q 
2 60 X 2 


= 27 - 


Let us now calculate the length of the chord c f which subtends 
the half of the arc occupied by one tooth. We have the formula 


and 

a = 

One will have, therefore, 


c = 2 r sin | a 
360° 


- = x ° 3 °'* 


4 X 60 

c = 2 X 54° X sin o° 45' 


log : 1080 = 3.0334238 
log sin o° 45' = 8.1169262 

log : (1080 sin o° 45') = 1.1503500 


Consequently, one will have 

Chord of one-quarter of the pitch = 14.1367 mm.* 

Let us lay off this length of / on a (Fig. 75) and draw the radius 
o a prolonged to the point b belonging to the epicycloid of the tooth; 
o b is then the total radius of the wheel and a b the height of the 
ogive of the tooth. 

On measuring a b, we will find it equal to 25 mm. and we will 
have the total radius of the wheel expressed in units of diametrical 
pitch, by the sum 

60 + ~ "LT 5 = 60 + 2 -77 = 62.77. 

One will place, therefore, the total diameter of the wheel at the 
division 62.77 of the compass, so that its primitive radius corres¬ 
ponds to the division 60. 

2 ^ 9 . One could proceed in an analogous manner for the pinion. 
Let us remark, however, that while the height of the ogive of the 
wheel is fixed, since it is formed by an epicycloid described by 
a point of a generating circumference with given radius, the excess 
of the pinion leaf is not so easily determined. 


* One could have obtained this result without the aid of trigonometry, by noting that 
the arcs and chords of small angles differ very little from each other. 

One would thus have 

= 4.5 X 3.1416 — 14.1372, 

4 X 60 


a very close approximation. 







Gearings. 


*93 


tt n ' 

n' 


TT. 


If this pitch comprises a third 
for the full tooth and two-thirds for 
the space, the length of the arc cor¬ 
responding to the thickness of one 
leaf being double the radius 8 of the 
circle of the excess, one will have 

8 = 


IT 



The form of the excess which one finds in a very great num¬ 
ber of pinions is that of a semi-circumference with radius equal to 
half the breadth of a leaf measured on the primitive circumference. 
This form, although we know it to 

be bad, especially for pinions of low . 

numbers, offers, however, a ready 
means for the calculation. 

Suppose n’ to be the number of 
leaves in the pinion. 

The primitive diameter expressed 
in function of the diametrical pitch 
will be likewise n\ since it is divided 
into as many equal parts as the pinion 
has leaves. 

The primitive circumference is, 
therefore, 

circumference = tr n' 
and the pitch of the gearing 


j —* 


• • 
» » 


rO 

Fie. 75 


2X3 


There must, therefore, be added a value equal to 8 to the two 

extremities of the primitive diameter of the pinion expressed in 

units of diametrical pitch : 

Total radius = n' + = n f -f- 1.05. 

3 


Thus, for the pinions whose full part of the pitch is equal to 
half of the space and whose excess has the form of a semi-circle, 
the total diameter should be stopped at the division corresponding 
to the number of leaves increased by 1.05. 

For the gearing which we will consider, of a wheel with 60 teeth 
and a pinion of 6 leaves, one should place the wheel at the division 
62.77 and the pinion at the division 7*°5* 








194 


Lessons in Horology. 


280 . Let us again make the calculation for a pinion in which 

the leaf is two-fifths of the pitch (12 leaves and above). 

As in the preceding case, the pitch of the gearing is equal to m 

and the radius of the ogive ir 

5 = —; 

5 

as 2 8 must be added to the primitive diameter, we will have 


The total diameter expressed in units of diametrical pitch is, therefore, 

11 ' + 1.25. 

281 . In the case of pinions with the excess semi-elliptical, the 
height of this excess becomes superior to those with which the pre¬ 
ceding calculations have furnished us. 

We have seen that the calculation of this value is complicated 
(259 and the succeeding) ; therefore, without entering into other 
details, we refer, for these values, to the table which we give further 
on (283, third column). 

Thus, taking up again our gearing of a 6o-tooth wheel and 
6-leaf pinion, we find in this table, that the total diameter of the 
pinion expressed in units of diametrical pitch is 7.4648 ; that is to 
say, 7.5 in round numbers. 

282 . Practically, to employ in a proper manner the proportional 
compass, one must, therefore, commence by examining the excess of 
the pinion leaves, estimating it with relation to the breadth of the leaf. 

If one judges, for example, that it is equal to half the thick¬ 
ness of the leaf, one will add a unit to the number of leaves n'; if 
the excess appears to be three-quarters of the thickness, one will 
add 1.5 and, finally, if the height is judged equal to twice the 
thickness of the leaf, one will add 2. 

A compound microscope, the eyeglass provided with spider 
lines and mounted on its lower side on a carriage furnished with a 
micrometer screw, allowing the object observed to move in the field 
of the instrument, can measure with great precision the height of 
the ogives of the wheels or of the excess of pinions. In default of 
this instrument, the method which we have just indicated is exact 
enough to be used. 

283 . The table which we give hereafter indicates the number 
of the division on the compass for the gearings most used in 
horology. Thus, on placing an 8o-tooth wheel at the division 




195 


Gearings. 


TABLE FOR USING THE PROPORTIONAL COMPASS 




DIVISION OF THE COMPASS 

THE RADIUS OR DIAMETER = 1 


NUMBER 

FOR SHAPE OF TEETH 

SHAPE 

OF TEETH 

DESIGNATION 

OF 






TEETH 

Elliptical 

Circular 

Elliptical 



Circular 

Wheel . . 

180 

183.542 


I.OI9676 




Pinion . . 

12 

13.66 

13-25 

1.14 

1.104 

Wheel . . 

144 

147.446 


I.024 




Pinion . . 

10 

n -5 

II.05 

I -15 

1.105 

Wheel . . 

96 

99-747 


I.03904 




Pinion . . 

12 

13.66 

13.25 

1.14 

1.104 

Wheel . . 

80 

83-3853 


I.0423 




Pinion . . 

IO 

11 .5 

II.05 

I -15 

1.105 

Wheel . . 

64 

67.1 


I.048475 




Pinion . . 

8 

9-45 

9-°5 

1.18 

I -13 

Wheel . . 

90 

93 - 6 i 4 


I.04016 




Pinion . . 

12 

13.66 

13.25 

1.14 

1.104 

Wheel . . 

75 

78.375 


I.045 




Pinion . . 

IO 

n -5 

11.05 

I -15 

1.105 

Wheel . . 

60 

63.0976 


1.05H 




Pinion . . 

8 

9-5 

9-°5 

1.18 

I-I3 

Wheel . . 

80 

83.1247 


I.039 




Pinion . . 

8 

9 5 

9-°5 

1.18 

I.I 3 

Wheel . . 

60 

62.7839 


I.0464 




Pinion . . 

6 

7.4648 

7-05 

1.2441 

I-175 

Wheel . . 

70 

72.9637 

— 

I.0423 

1.15 real diameter 

Pinion . . 

7 

8-397 

8.05 

I-1995 

1.1 on pressing 2 
leaves on one side 

( < 

7 

7.972 

7-7 

I-139 

and 1 on the other 

Wheel . . 

48 

5o-77 


1-0577 




Pinion . . 

6 

7-4 

7-05 

1.23 

1.175 

Wheel . . 

06 

38.74 


1.0762 




Pinion . . 

6 

7-4 

7-05 

1.23 

I-175 

Wheel . . 

30 

32.72 


1.090S 




Pinion . . 

6 

7-4 

7-°5 

1.23 

1.175 

Wheel . . 

36 

38 55 







Pinion . . 

12 

14.02 




Wheel . . 
Pinion . . 

40 

40.7 

11.52 







IO 













































































































































































196 


Lessons in Horology. 


83.38 of the compass, a 10-leaf pinion in which it should gear 
should correspond to the division 11.5 if the excess is of semi¬ 
elliptical shape, or at the division 11.05 if this form is semi-circular. 

For a pinion with the uneven number of 7 leaves, one will 
find two indications in the table, one giving the real diameter, the 
other permitting the placing of the pinion with one leaf pressing 
against an arm of the compass, and the two leaves opposite against 
the other arm. This last measure comprises, therefore, in units of 
diametrical pitch, a total radius increased by the versed sine O B 
(Plate II). 

After what we have said, it will be easy to obtain in a graphical 
manner the figures corresponding to gearings not appearing in 
the table. 

284. Verification of a Proportional Compass. The two divided 
scales should be perfectly straight and consequently in exact juxta¬ 
position when the instrument is closed ; this, one verifies by hold¬ 
ing the instrument to the height of the eyes and seeing if the two 
scales are perfectly fitted against each other. 

The divisions should be regular and the zero point should be 
found in the center of the hinge. 

It is also easy to verify this condition with exactitude by taking 
off, with a pair of sharp-pointed dividers, a certain number of 
divisions, 10, for example : on moving, then, these dividers over 
the whole of the part divided, it is easy to assure oneself of the exact¬ 
ness of this condition. Finally, on placing one of the points of the 
dividers on the division 10, one should be able to place the other 
on the center of the hinge. 

This hinge should be made in such a manner that the arms 
can be spread without any jerk, that is to say, with even friction ; in 
no case could any play or shake whatever be allowed at this hinge. 

A compass being thus verified, it could be used with the aid of 
the given table. 

There exist other systems of proportional compasses, most of 
which dispense with the use of an accessory table. Let us remark, 
however, that the one which we have just described has its principle 
founded on an exact and rational basis and that the table which it 
requires complicates its use very little, if at all. 

285. Determination of the Distance Between the Centers of a 
Gearing by Means of the Proportional Compass and of a Depthing 
Tool. Having fastened the proportional compass in such a manner 


Gearings. 


197 


that the primitive radius of the wheel corresponds to the figure 
for its number of teeth, one measures, in this same opening, the 
diameter of one of the arms of a depthing tool. Let d be the 
division corresponding to this last 
measure. One opens, then, the 
depthing tool until the two arms a 
and b, drawn in section (Fig. 76), 
correspond to the division 

n -f n' 

-b d. 

2 

This opening then gives the distance 
between centers. 

Example : Having regulated the 
opening in the proportional compass 
so that the total diameter of a 60- 
tooth wheel is fitted to the division 
62.78 (see the table), one measures 
the arm of a depthing tool and finds 
that its diameter corresponds to the 
division 8 ; we will thus have 


60 -f- 6 


+ 8 — 41, 



Fiir. 70 


the pinion having 6 leaves. The 
opening of the depthing tool should 
then be regulated in such a manner 
that the two arms a and b corres¬ 
pond to the division 41. 

286. The Proportional Compass and Stem-winding Gearings. 

First—Gearing of the crown wheel in the ratchet wheel : The teeth 
of this gearing should be solid ; this is the reason why only one- 
twentieth of play is given them (223). For the same purpose the 
bottoms of the teeth are made with a rounded shape and the ogives 
of the teeth are shortened. These gearings are epicycloidal; the 
profiles are formed by epicycloids described by a point of a gene¬ 
rating circumference smaller than half of the primitive circumfer¬ 
ences. As we have just said, one does not use the whole of the 
epicycloidal arc for the tooth ; it is sufficient that the contact be 
established three-fifths of the pitch before the line of centers, in 
order to be continued until three-fifths of the pitch beyond that line. 





198 


Lessons in Horology. 


The “flank” of the tooth is no longer a straight line, but a 
hypocycloid described by a point of the same generating circle 
rolling on the interior of the primitive circumference of the wheel. 

Thus (Fig. 77) t a is the useful epicycloidal arc, while a b is 
any curve whatever shortening the tooth ; in this manner, the height 
of the ogive is not determined ; t d is a hypocycloidal arc gen¬ 
erally approaching, very nearly, a straight line. 



To determine the height of the shortened ogive in units of 
diametrical pitch, it is necessary to proceed graphically or by simply 
estimating it by the eye 

Generally, for this gearing, the height of the two ogives placed 
opposite to each other can be taken as 2^ diametrical pitch. 

If, then, n and n' are the numbers of teeth, the crown v/heel 

should be fitted to the division 

n' 4- 2]i 

and in the same manner the ratchet to the division 

n + 2%. 







199 


Gearings. 


Let us note, however, that since the crown wheel always drives 
the ratchet, it is preferable to make the first proportionally greater 
than the second ; for example, 

Crown wheel, division . ... n' 2}% 

Ratchet wheel, division . . . n -f 2% 

28?. Gearing of the Winding Pinion in the Crown Teeth of the 
Contrate Wheel. In these gearings the axes of the two mobiles 
form a right angle between them. Logically, such a gearing 
should be a conical gearing (311) ; in the practice of horology 
it is sufficient, however, to skillfully simulate it. 

One finds two general 
arrangements of this system, j- 

In the first (Fig. 78), 
a b is the exterior diameter 
of the crown teeth in the 
contrate wheel ; this is, at 
the same time, its primitive 
diameter, for the ogive of 
the tooth is not to be added 
to the extremity of the radius, since the teeth are perpendicular to 
the plane of the wheel. The crown wheel must, therefore, be fitted 
to the division n* of its teeth and the total diameter c d of the 
pinion, perpendicular at b on a b to the division n -f- 2^, as in the 
preceding case and also for the same reason. Therefore, 

Winding pinion, division. n -f- 2^ 

Crown teeth of the contrate wheel, division . . n' 



The second arrangement is found in some winding mechanism. 

It admits of a teeth 
range with the crown 
teeth outside of its 
primitive radius (Fig. 
79); in this case there 
must be added to each 
of the two mobiles 


the height of the two 



Fig. 79 ogives. One will then 

have 

Winding pinion, division . . . n -f- 2^ 

Crown wheel, division . ... n f -f- 2 

















200 


Lessons in Horology . 


288. Gearing of the Sliding Pinion and of the Small Setting 
Wheel. Although one could not make use of the proportional 
compass for the study of the relative dimensions to be given to the 
mobiles of this gearing, and as this determination should be 
entirely a matter of calculation, we give here, however, the theory, 
which will not be found out of place. 

Suppose (Fig. 80) r to be the primitive radius of the small 
setting wheel, R its total radius and n the number of its teeth ; 



r' the radius of the sliding pinion abutting on the under side of the 
small winding wheel, R’ its total radius and v! the number of 
its teeth. 

In the generality of cases, one can admit that 

. R / — r' — 0.2 mm.: 

consequently, 

r' = R' — o.2. 


If the penetration of the two mobiles is greater, the gearing 
does not work well. Generally, it is desired in practice to deter¬ 
mine the number of teeth n' in the sliding pinion. 

The proportion r n 

r' n' 

gives us the value , 

(i) n f — n —. 

V r 


By an analogous reasoning to that of (286) one can place 

r _ n 

R n -f- 2 


R n 


from whence one draws 





















Gear bigs. 


201 


Replacing 

becomes 


r and r 1 by their value in the equation (i), it 
,/ — 71 ( R ' ~ °- 2 ) 


n' 


or 


R n 
n -j- 2 


n' 


( R' — 0 . 2 ) (n 4 - 2 ) 
r 


gives 


(2) 

Example : Let R — 2.4, R' — 2 and n = 18, the formula (2) 
(2 — 0 . 2 ) (18 + 2 ) 1.8 X 20 


n' — 


15 teeth. 


2.4 2.4 

289. Gearings of the Dial Wheels. The small and the large 
setting ivheels are placed in the proportional compass at the division 
corresponding to their number of teeth increased by 2. Therefore, 

Small setting wheel, division . ... n -f 2 
Large “ “ “ + 2. 

For the gearing of the cannon pinion and the minute wheel , 
it is generally the same ; however, notice must always be taken of 
the form which the excess has, in the leaves of the cannon pinion. 
If the leaves are terminated by a semi-circular form, it would then 
be necessary, in this case, to place the pinion at the division corres¬ 
ponding to the number of leaves increased by 1. Sometimes, also, 
the teeth of the minute wheel are formed in such a manner that one 
is obliged to add 2.5 or even 3 to their number, in order to obtain 
the division of the compass at which this wheel should be placed. 

In order to verify at one time the series of relations between 
the wheels for setting the watch, the cannon pinion must be fitted 
to the division corresponding to its number of leaves increased by 
2, and the other wheels, dial wheels and large and small setting 
wheels, should likewise correspond to the divisions for their respec¬ 
tive number of teeth increased by 2. 

If there should be a fault, it is always better that the wheel 
which drives be a trifle large. Since one prefers, in this train, an 
easy and smooth transmission, and since the small setting wheel is 
the wheel which drives, one can allow it to be slightly larger than 
a strict proportion would give it. 

The gearing of the minute wheel pinion and the hour wheel is sub¬ 
mitted to the same law with the same reserves for the different forms 
of excess which are found in practice. Generally, however, one has 

Minute wheel pinion, division . ... n -f- 2 


Hour wheel, division, 


u! 4 - 2 . 







202 Lessons in Horology. 

Various Calculations Relative to Gearings, 

290 . The preceding table (283) gives a second factor for the 
various gearings which it includes, expressing the radius or total 
diameter of the mobile in function of the radius or primitive 
diameter equal to the unit. The use of these factors is impor¬ 
tant in cases where it is desired to determine the total dimensions 
of the wheels of a gearing by means of a rapid calculation ; for 
example, in the construction of calibres. The solving of the fol¬ 
lowing problems will rapidly render us familiar with the use of 
this table : 

291 . Bemg given the primitive radius of a wheel to calculate 
its total radius R. To solve this question, the number of wheel 
teeth and of pinion leaves in which they should gear must be 
known. One seeks, therefore, in the second column of the table 
for the figures of the teeth range of the gearing, and the corres¬ 
ponding value indicated in the fifth column will give the factor by 
which the primitive radius of the wheel must be multiplied to obtain 
its total radius. One will, therefore, have 

R — r X factor of the table. 

If, for example, the primitive radius r of a barrel with 80 teeth 
is 10 mm. and if this wheel ought to gear in a 10-leaf center 
pinion, the factor indicated by the table is 1.0423 ; therefore, 

R = 10 X 1-0423 = 10.423 mm. 

If we had to calculate the total radius of a barrel with 96 
teeth, whose primitive radius should likewise be 10 mm. and gear¬ 
ing in a 12-leaf pinion, we would have 

R ~ 10 X 1*03904 = 10.3904 mm. 

One sees that the total radius of the latter barrel is a little less 
than that of the first. 

292 . Being given the total radius of a wheel , to determine its 
primitive radius. This question is the inverse of the preceding 
one and consequently is solved by dividing the total radius given 
by the tabulated factor. Therefore, 

R 

- - -- - ^ 

tabulated factor' 

Numerically, one has, if R = 10.423, n — 80 and n' = 10, 

__ 10.423 


10. 




Gearings. 


203 


293 . Being given the primitive radius of a pinion, to calcu¬ 
late its total radius and, reciprocally, being given the total radius 
of a pinion, to find its primitive radius. The same as for the 
wheel ; in the first case, one multiplies the primitive radius given 
by the factor of the table, and in the second, one divides the total 
radius by the factor. One has, therefore, 


or 


R' = r' X tabulated factor, 
R' 


tabulated factor’ 

Just as for the proportional compass the table gives two factors 
for each pinion, one suitable for a semi-elliptical excess, the other 
for a semi-circular excess. 

Let, for example, r' = 1.25 and n’ = 10, semi-circular form ; 
one will have 

R' = 1.25 X 1.105 = 1.38025, 


or 


r' 


i- 38o 2 5 
1.105 


= 1-25. 


If the excess was of semi-elliptical shape, one would have 

R / = 125 X 1.15 = 1 - 4375 . 
or, for the inverse problem, 

✓ = i«75 = 


I-I5 


294 . Being given the primitive radius of a wheel, one seeks 
for the total radius of the pinion in which it gears (semi-elliptical 
form). One has the proportion 


n 

n' 


r* — r 


n' 

n 


which gives the value 
and since 

R / = yf x tabulated factor 
one obtains, on replacing r' by its value, 


R' = r 


n' 

n 


X tabulated factor. 


Thus, as example, 

r — 5.38, n — 75, n' = 10 


R' = 5-38 X^X 1.115 = 0 8249. 

/ o 


one will have 








204 


Lessons in Horology. 


295 . Being given the primitive radius of a pinion , one desires 
the total radius of the wheel. We have 


r — r' - 


n 


n 


7 » 


and 




= r X tab. fac. = X tab. fac. 




Let X = o.86, = 8o, n' = io. One will write 


=• o.S6 X 


8o 


io 


X 1.0423 = 7.17- 


296 . Being given the total radius of the wheel , to find the total 
radius of the pinion. We have 

R 


r = 


and 

on replacing, it becomes 


tab. fac. of the wheel 
n' . 


r = r — 








7 ? 


Afterward 
consequently, 

Let, for example, 


n tab. fac. of the wheel 
R' = r' tab. fac. of the pinion ; 

r n' tab. fac. of the pinion 
n tab. fac. of the wheel ’ 

R = 10.2, n = 80, n' = 10, 


one will have 


Tabulated factors j I -° 4 2 3 

{ rmion = 1.15 


r ' = 10.2 x-^x = 1.4067. 


80 ' ' 1.0423 

29 T. Being given the total radius of the pinion , to find the 
total radius of the wheel. In an analogous manner to the preced¬ 
ing case, one will have 

X 

and 

from whence 


_ Rf _ 

tab. fac. of the pinion 

, n 


7V 


R' 


n 


And since 
one has at last 


tab. fac. of the pinion n' 

R — r tab. fac. of the wheel, 

n tab. fac. of the wheel 
n' tab. fac. of the pinion * 


R = R' 













Gearings. 


205 


Let R' — 1.4067, n — 80, n' — 10, one will have 

r — I -4°^7 X 8 0 X 1.0423 


= 10.2. 


10 X 1.15 

298 . Being given the distance of the centers and the numbers 
of teeth in a gearing , to determine the total diameters of the wheel 
and pinion. We know the formulas (185) 


r — D 


and since 
one has 


n 


71 A V n 71 

-, > and r 7 = D - ; - 

+ n' n + n 


R — r tab. fac. of the wheel, 
R' — r' tab. fac. of the pinion, 


R 


D 


n 


n -f n' 


tab. fac. of the wheel, 


R' = D 


n' 


Let 


7 tab. fac. of the pinion. 

: 7 , 


n 4- n 
D = 5.2, n — 70, n' 


one will have 


Tabulated factors { I ‘° 4 2 3 | or wheel 

[ 1.1995 for the pinior 


pinion 

2 R = — X 70 X X 4 ' 3 X 2 = 9 - 854 , 

2 R' = 5 - 2 X 7 X 1-1995 X 2 = , 

77 

299 . Being given the total radius of the wheel and the num¬ 
bers of teeth of the gearing , one desires to find the distance between 
the centers of the two mobiles. The formula 

n 


gives 
and since 
one will have 


r ■= D 

D = r 


n 4- » 
n 4 ~ n' 

y 

n 

R 


r = 


D 


tab. fac. wheel 
R n 4 - n' 


Suppose 


One places 


tab. fac. wheel n 

R — 4.927, n = 70, n' = 7, 
Tabulated factors of the wheel, = 1.0423. 

4927 X 77 = 

1.0423 X 70 5 ' ' 


300 . Being given the total radius R' of a pinion and the num¬ 
bers n and n' of the teeth of the gearing , to determine the distance 
of the centers D. 















206 


Lessons in Horology. 


In an analogous manner to the preceding 

write the formula _ D , , , 

D — R n n' 

tab. fac. pinion 


Suppose 


n' 


R' = 0.567, n — 70, n' 


7, 


one w r ill have 


D 


0-5 67 X 77 __ 2 

1.1995 X 7 


case, we would 


301 . Being given the diameter P of a watch plate , the num¬ 
bers of teeth n of the barrel and n! of the center pinion, one desires 
to fi 7 id: 1st, the primitive radii r a?id X of the wheel and of the 

pinion; 2d, the dista?ice of the centers D of the two mobiles; 

jd, the total radii R a?id R'. The diameter of the barrel should 
be as large as possible. This question is generally one of the first 
which presents itself in connection with the establishing of a new 
watch calibre. 

In order to be able to fit the plate of a watch in its case, a 
“recess” is generally made on the exterior of this plate, which 
can be valued at one-sixtieth part of the total diameter P. 

There remains, therefore, only available. The useful radius 

is, consequently, 


V 59 
* 60 


P = 


59 p 
120 


The extremity of the teeth range in the barrel can coincide 
with the extremity of this radius, the teeth finding the necessary 
play in the hollowed-out part in the center of the case. 

The radius T 5 y 9 y P should be equal to the sum of the following 
lengths : the primitive radius r' of the center pinion, the primitive 
radius of the barrel and its total radius (Fig. 81). 

One would, therefore, have 


But since 
and 


59 

120 


P — p + r -j- R. 


i'/ 


71' 

n 


R = r tab. fac., 


one will also have sq 11' 

P — r •- |- r r tab. fac., 


120 


71 


or, again, 


59 

120 


P 


= r -fi 1 + tab. fac. ^ 








Gearings. 


207 


from whence one finds 


r = 


59 

120 


P 


n' 

n 


+ 1 -f tab. fac. 


Knowing r, it is easy to determine r' } since 


r f — r 


n' 

n 


the distance of the centers D will afterward be determined by the 
sum of the two primitive radii. 



Knowing r and r', one will calculate the total radii R and R' 
by the operation : 

R — r tab. fac. of the wheel. 

R' = 7 ' tab. fac. of the pinion. 


Suppose, as an example, P — 43 mm. (19 lines), 71 — 80 and 
n’ =10. 

Tabulated factor of the wheel = 1.0423. 

Tabulated factor of the pinion, = 1.15. 

One will have : 


59 

120 


X 43 


r 


1+1+ 1-0423 














208 


Lessons in Horology. 


or also 
and 

It follows that 
Then 

and finally 
Since one has 


_59 X 43_ = 59 X 43 

120 (i + i + 1.0423) 260.086 

r = 9-7545- 


, _ 9-7545 
8 


1.2193. 


D = 9-7545 + 1.2193 = 10.9738, 


R = 9-7545 X 1.0423 = 10.17. 
R! = 1.2193 X 1.15 = i- 4 - 

S' « = 


one should have likewise, or very nearly, 

r* 4 ~ r + R = 21.1416. 

The addition gives 

1.2193 

9-7545 

10.17 

21.1438. 


With this approximation the result is satisfactory. 

302 . Being given the total radius R' of the pinion in which 
the rack of a repeater gears , the 7 iumber n•/ of teeth according to 
which the pinion is divided * and the total radius R of the rack , 
one desires to find the number according to which the sector of this 
last wheel must be cut. 

Let us admit the ogive of the teeth range of the two mobiles 
equal to twice the diametrical pitch (277). We will have the 
primitive radius of the pinion by the formula 


R' n' 


and that of the wheel 

n f 4- 2 


R n 

r — ■ -:—• 

Since one has 

« 4* 2 


r* n' 

one can also write 

r n 


R' n f 

n' 

n' 4-2 R' 

n 

R n ~ R 


R' n' (n 4 “ 2) 
R n {n' 4- 2)’ 


n 4 - 2 


*0ne knows that the division of this pinion by the set of teeth is not complete, for the 
reason that this mobile only executes a fraction of a turn. The pitch of the gearing left full 
facilitates the arrest of the movement. 


















Gearings. 


209 


from whence R „ {n , + a) = „ R , n , {n + 2) 

and, on simplifying, 

R (»' + 2) = R' (n -f 2). 

One finally obtains 

R , , . . 

« = -g, (»' + 2) — 2. 


Suppose, for example, R — 9.96, 7 ?' = 
would have : 

- ? - 96 - X 15 -2 = 81 teeth 


1.8, 71 ' = 13, we 


n 


1.8 


Remark.— One could obtain directly the above formula by a 
proportion analogous to that of 289, 

R n - J- 2 

R' n' -f- 2 


303. The following problem does not find its solution in the 
theory of gearings only, but also in that of trains and of the 
motive force. It recalls to our mind, in an excellent manner, the 
studies that we have gone over, so we do not hesitate to close this 
series of problems by joining together some of the various ques¬ 
tions which we have treated in this chapter and in those which have 
preceded it. 

304. Problem. A horologist has constructed a stem-winding 
watch the diameter of whose plate is 50 millimeters (22 lines). 
Upon winding the watch, he notices that the power necessary to 
operate the winding works, that is to say, to overcome the force 
opposed by the spring to the movement, is too great. He decides 
then to manufacture a new watch, like the first, but in which the 
winding can be more easily effected. He should, therefore, modify 
the value of the two factors which enter into the expression of the 
mechanical work : the force on the one hand and the tune employed 
for the winding on the other ; in other words, the space traversed 
by the point of application of the active force (37). 

The first watch has its barrel furnished with a stop work of 
4 turns, and runs for 32 hours ; the second should run for the same 
number of hours. It is evident that if we introduce into the 
second watch a barrel furnished with a stop work, allowing it to 
make 5 rotations during 32 hours, we will have diminished the 
average tension of the force and augmented at the same time the 
duration of the winding. 





210 


Lessons in Horology. 


Let us seek, therefore, for the nature and the value of the 
change that must be made in the second construction, in order to 
arrive at the end desired. 

305 . In the first place, the relation between the numbers of teeth 
in the barrel and of leaves in the center pinion must be changed. 

In the first watch, this relation was ff, which gave a length of 
running 


96 

12 


X 4 = 32 hours. 


In the second watch, one should also have 


n 


or 


n 


7 X 5 = 32 hours, 


* = - 6.4. 


n- 5 

On choosing a pinion of 10 leaves, one will have 

n = 6.4 X 10 = 64 teeth, 
or a pinion of 15 leaves, 

« = 6.4 X 15 = 96 teeth. 

It is not an absolute necessity, in general, to conserve the 
above relation in a very strict manner. Thus, if one wished a 
pinion with 12 leaves, one would have 

n = 6.4 X 12 = 76.8 teeth. 


This fractional number not being practical, let us admit, for 
example, 

We would then have 

78 


12 


71 = 78. 

X 5 == 32.5 hours. 


The watch would run, with the above number, half an hour 
longer than was desired. 

Since, for a watch of 50 mm. diameter, a barrel with 96 teeth 
does not give too weak a teeth range, one can accept for this 
gearing 

96 teeth for the barrel, 

15 leaves for the center pinion. 

306 . The primitive radii of the two mobiles must now be cal¬ 
culated and the distance between their centers. Let us commence 
by seeking for these values in the first watch, in order to compare 
the results. 






Gear mgs. 


211 


The formula which we have determined (301) gives us 

59 

r = _ 


120 


X 50 


12 

96 


+ I + I.O39 


and making the calculation r _ x g 


Then 


n' 


12 


and 


r r T = "- 36 x ^6 = J ' 42 ' 


D — 11.36 -f 1.42 -= 12.78. 
The total radii of the two mobiles were : 


R — 11.36 X 1.039 == 11.803, 

R’ — 1.4 2 X 1.14 = 1.619. 

For the second watch, the calculations are naturally analogous, 
only we have here the case of a gearing whose teeth range is not 
found in the table. 

One can admit, in this case, by analogy the tabulated factor of 
the wheel equal to 1.04 by slightly forcing the figure of the factor 
in the preceding gearing, since the height of the ogive should 
increase with a larger pinion (15 leaves instead of 12). 

For the pinion, we will concede an excess with semi-circular 
shape perfectly admissible for this number of leaves. 

We will thus have 


59 

120 


X 50 


45 

96 


+ 1 + 1-04 

and making the calculation 

& r = 11.17 ; 

and , 15 

r' = 11.17 

Therefore, 


96 


= 1 - 745 - 


D = 11.17 -f 1.745 


12.915 


one will have the total radii 


and 


R = 11.17 X 1.04 = 11.617 


R' = r' + 


2 it r' 
15 X 5 



One sees that the diameter of the barrel has diminished and that 
of the pinion and the distance between the centers has increased. 

30 <T. Let us now seek for the exterior and interior radii of the 
barrel drum. 

These dimensions ought to be as large as possible. The ex¬ 
terior radius of the drum in the first watch was 11 mm., therefore 









212 Lessons in Horology. 

0.36 less than the primitive radius. That of the second watch 
could, therefore, be 

11.17 — 0.36 = 10.81 mm. 

The interior radius of the drum was, in the first case, 

11 — 0.77 = 10.23; 

consequently, it could be, in the second case, 

10.81 — 0.77 = 10.04 or 10 mm. 

308 . Let us now calculate the dimensions of the hub, and the 
dimensions as well as the force of the spring. 

In the first watch, when the spring is pressed against the side 
of the drum, it occupies the third of the interior radius of the 
barrel, another third remains empty and the third third is occupied 
by the hub. This spring makes, in this position, 15 turns and 4.5, 
placed loosely on a table ; the number of turns of development is 6.5. 
The dimensions of the blade are the following : 

Thickness = 0.21 
Height =3.9 
Length = 780 mm. 

According to the formula (97) 

E h e 3 2 ir n 


the moment of the force of this spring is for n = 1, 

F = 557 - 64 - 

When the spring is coiled up, that is to say, when the watch 
is completely wound, one has 

n = 15 + 6 — 4.5 = 16.5; 
consequently, F — 16.5 X 557 = 9190 gr. 

When the spring is down, one has n' = 12.5 and 

F' = 12.5 X 557 = 6942 gr. 

In order that the center wheel may receive in the second watch 
the same force at the beginning and at the end of the tension of the 
spring, it is necessary that, the watch being wound, one should have 

F\' = X 9190 = 7352 gr., 

and if the watch is run down 

4 


Fx’ = 


5 


X 6942 = 5553 gr., 





Gearings. 


213 


since for the same number of teeth in the barrels the relation of the 
leaves and pinions is I2 4 

15 ~ 5 ’ 


In order that this difference between the moments of extreme 
force may exist with 5 turns of the stop work, one must have the 
proportion _ »/+ 5 

12.5 _ 7 lg ' 

from whence one finds , 

ng = 15.625, 


and afterwards 


n x = 20.625 — n. 


309 . Let us further calculate the thickness of the spring. Let e 
be this thickness for the spring in the first watch and e' that of the 
spring in the second. 

For the first case we have the moment 


and for the second 


5 


F = 


F = 


E h e* 2 it n 
12 L 

E h e n 2 7T | n 

T e * 

12 L. - 


on remarking that the length L increases in inverse relation to the 
diminution of the thickness e. 

One could, therefore, place 


4 E h e* 2 it n _ E h e'* 2 ir \n 

T ' 12 L ~~ I2 s 

from whence, on simplifying, one obtains 


and 

therefore, 
from whence 



For e — 0.21, one has 

^ = 0.21 X 0.894427 = 0.10826. 

310 . Since we have the proportion 

L e' 




















214 Lessons in Horology. 

one could finally calculate the new length of the spring on placing 

L' = L ; 

6 ' 

therefore, 

U = —~ - - 780, 

0.894427 

and 

IJ = 872 mm. 

Since the second barrel is a little smaller than the first, this 
spring will fill it a little more than one-third ; but as it is also 
thinner, one can diminish the hub proportionately to the relative 
thickness of the spring (m). 

Thus, the first hub having a radius equal to one-third of the 
interior radius of the barrel, this one would be = 3.66 . . mm. 

This radius being 17.777 times greater than the thickness of 
the spring, the second hub could be 

17-777 X .18826 = 3.34 . . . mm. 

311 . We know, moreover, the means of increasing the ease of 
the winding by increasing the number of teeth in the ratchet and of 
the contrate teeth range in the crown wheel ; since we have already 
treated this question (169), we will not go back to it here and we 
will thus admit that the problem proposed is solved. 

Conical Gearings. 

312 . In the gearings that we have just considered, the two axes 
are parallel to each other and we know that the movement of the 
system can be compared to that of two cylinders mutually conduct¬ 
ing each other by simple contact. We have designated gearings of 
this sort under the name of cylindrical gearmgs. 

313 . If, in place of being parallel, the two axes are concurrent, 
one can imagine that the movement of one produces the movement 
of the other by the contact of two cones concentric to each of the 
two axes (Fig. 82). This system takes, therefore, the name of 
conical gearing. 

The two axes can form any angle whatever with each other ; 
we will treat particularly the special case where this angle is a right 
angle, almost the only case in horology. 

Suppose (Fig. 82) O x and O y, two perpendicular axes 
around which turn the two cones COB and A O C; let us admit 
that their movement is produced without slipping. 



Gearings. 


215 


As for the cylindrical gearings, one can prove that the rela¬ 
tion of the angles traversed by the two cones is inversely propor¬ 
tional to that of the diameters C B and A C. 

The diameters can be measured in any manner whatever, pro¬ 
vided that their circumferences be tangent. Thus, in place of C B 
and A C one can just as well take C' B' and A' C', since these 
straight lines form the sides of similar triangles. 



a and a ' being the angles traversed in the same time by the 
two cones, one has, therefore, the proportion 

A C _ A' C' = 0/ 

C B ~ C' B' 'a ’ 


and since, n and n ' being the numbers of teeth 


one will also have 




} 


A C __ n 
CB ~ n r 


314. The pitch p of the gearing varies with the distance O C; 
for such a point of contact C, it is 


A C 


B C 
n' 


p — IT 


n 


IT 














2l6 


Wessons in Horology. 


315. Form Of the Teeth. As in the cylindrical gearings, the 
transmission of the movement cannot be effected practically by the 
simple contact of the two primitive cones ; one is generally obliged 
to supply these cones with flanges, that is to say, with teeth, which 
make them move as if they were driven by their simple adhesion. 


\ 

\ 



The contact, consequently, is not always found on the line O C ; 
if we consider the contact at the point C } the displacement of this 
contact should take place on the surface of a sphere whose center 
is at O and passing through the points A C B. The form of the 
teeth must be traced on this sphere. 

Thus, M O and O iV(Fig. 83 ) being the two axes of rotation 
which meet in a point O , let us take this point as center of a 
sphere ; it will contain the two upright cones, as we have just said, 
having their common summit at the center O , and will cut them 
along two circumferences, of their bases, tangent at the point A 
belonging to the generatrix of contact of the two cones. 

These circumferences drive each other, exactly as would the 
primitive circumferences of a cylindrical gearing situated in the 
same plane. The sphere playing the role of the plane considered 



Gearings. 217 

in the first case, all the properties already determined are here 
reproduced. 

Thus, one can determine the curve described by a point of 
one of these primitive circumferences by making one of the cones 
roll on the other which remains immovable ; the curve thus engen¬ 
dered is the spherical epicycloid B C. 

On account of the similarity of the methods employed in this 
case with those that we have previously described, we will not enter 
into all the details of these constructions. 

316. Let us examine, however, the case of flank gearings. 
The flank being a diametrical plane of the primitive cone, the driving 
tooth will be a conical surface whose form must be determined. 

Suppose S’ O and .S' O' the axes of the two primitive cones 
which should turn while touching along an edge 6 " M (not repre¬ 
sented in the Fig. 84 ). Suppose Mm iVand Mm' N’* the circum¬ 



ferences of the circles proceeding from the intersection of the two 
cones by the planes drawn perpendicularly through the point M to 
their respective axes. On the radius M O' of the circle O' as 
diameter, one describes a circumference 0" and through its center 







2 iS 


Lessons in Horology. 


O" one erects a perpendicular on its plane ; this perpendicular will 
meet the axis at a point 2 . 

If one considers this point 2 as the common summit of two 
cones having for bases the two circles O and O", and if one makes 
the second cone 2 O" roll on the first 2 O, a point of the circum¬ 
ference O" will describe a curve m m", a spherical epicycloid , 
situated on the sphere on which is moved the circle O" itself, 
sphere having its center at 2 . 

If one made a cone which had its summit at the point pass 
through this epicycloid, this cone will be the exterior surface of a 
diametrical plane of the cone O' and should consequently be 
taken for the surface of the cone .S' O. This result appears evident 
from the similarity in the construction of cylindrical gearings, there¬ 
fore, we will add no other 
proof to the application 
of this development by 
analogy. 

317. Besides the epi- 
cycloidal form, one em¬ 
ploys also the evolvent of 
circle for the teeth of coni¬ 
cal gearings. 

318. Construction of 
Conical Gearings. By the 
preceding, all the lines 
which enter into conical 
gearings being defined, it 
is only necessary to apply 
the principles of descrip¬ 
tive geometry to deduce 
from them the outlines nec¬ 
essary for its construction. 
But it is useless to enter 
into extended details with 
more simple method has 

been adopted and one sufficiently exact. 

In order to represent these forms in a more convenient 
manner on paper, one substitutes for the spherical surface the 
plane surface projected at F C G (Fig. 85 ) perpendicular to the 
line of tangency of the primitive cones O C. On this surface 



regards to this, since in practice a 






Gearbigs. 219 

one represents the developed surfaces of the two cones A G C 
and B F C (Fig. 86). 

In the development of the cone O D , the circumference pro¬ 
jected in ADC will become an incomplete circle A C A with 
radius A G of the same length as 
this circumference. Likewise, the 
cone O E developed will give an 
arc of circle of same length as the 
circumference projected in CEB. 

It is on these circumferences 
that one lays off the lengths cor¬ 
responding to the pitch of the 
gearing determined on the circum¬ 
ferences with radii A D and C E of 
the bases of the two cones. One 
draws afterward the form of the 
teeth of the two mobiles, as one 
does it on the primitive circum¬ 
ferences in the cylindrical gearings. 

These forms are obviously 
equal to those that one would ob¬ 
tain on the spherical surfaces them¬ 
selves, since, for the small dimensions of a tooth, the surface of 
the plane and that of the sphere are almost the same. 

For the purpose of being able to compare the form of the 
teeth of a conical gearing with that of the drawing, one terminates 
these wheels by a portion of the cones C G A and C B F (Fig.85). 

319 . In horology, however, one cannot do this, either on 
account of the lack of room or because the wheel carries at the 
same time another teeth range, as the crown wheel of stem-winding 
gearings, for example. The exterior surfaces of the two wheels 
are then straight planes, perpendicular to the axes. 

Admit, for this case, that the exterior planes of the two 
mobiles meet in C (Fig. 87), and let us seek for the form of the 
teeth cut by the plane C B. It is, in fact, on this plane that 
we see the form of the teeth range and that we can determine 
its dimensions. 

Let us first draw the two primitive cones COB and A O C. 
the latter being represented only by its half DOC ; draw the 
perpendicular F G to O C and a parallel F' G' to F G , finally the 






220 


Lessons in Horology. 


primitive circumferences with radii F' C' and G' C' and determine 
the form of the teeth according to the method known. 

In order to obtain a horizontal projection of the tooth of the 
wheel (winding pinion, for example) whose center we can place at 



O ', let us note that the point C’ is projected at C and C", we will 
have, therefore, the circumference of the primitive base of the cone 
with radius EC— O" C let us lay off half the thickness of the 
teeth C' H' on each side of C" on the primitive circle. 








Gearings. 


221 


In order to determine now the total radius O" /", correspond¬ 
ing to the point /', let us project this point /' on the plane F G 
at /; draw the radii O / prolonged to M ; the point of intersection 
M of this radius with the plane B C gives the total radius £ M 
that one can project on O" /"; from the center O" describe after¬ 
wards the total circumference of the wheel. 



Fit;. 88 


One proceeds in an analogous manner to determine the bottom 
of the teeth on the plane B C by projecting the point K ' on F G , 
drawing the radius O K cutting the plane B C at 7/and projecting 
this point at K "; one will have thus the radius O K" of the cir¬ 
cumference passing through the base of the teeth. 

In order to determine finally any point P" of the form of the 
teeth, project the point P' to P on F G, draw the radius O P U 
and project the points P U on O” /". From the center O" one 
causes to pass through the points obtained arcs of a circle and lays off 
the half thickness S' P' on the circle passing through S"] and 

















222 


Lessons in Horology. 


draws, afterwards, the radius O" T" . The point P" at which this 
radius just cuts the circle projected from U, is the point of the tooth. 

Let us still seek for the form of the teeth which appears on the 
interior plane L Q parallel to the plane B C M (Fig. 88 ). 

On the side elevation, the point Q represents the point of the 
teeth. Project this point on the front elevation, at Q", and describe 
from O" as center the circumference which passes through this point 
and which gives us the point of the teeth. To obtain the point 7 " 
of the base of the teeth, draw the radius O N, cutting the plane 
L Q at T and project this point on the front elevation, we will 
thus obtain the point T" through which one passes the circum¬ 
ference of the base of the teeth. 

In order to further obtain any points whatever, for example, 
those which are found on the primitive cone, one draws the radius 
O C , cutting the plane L Q at V, projects this point on the front 
elevation, at V" and describes the circumference from the center O". 
One afterwards draws the radius O" B ; the points determined by 
the intersection of this radius with the circumference passing 
through V", are points of the curve of the teeth. One can thus 
determine as many points as one desires and represent in this 
manner the complete form of the tooth. 

Let us remark that, compared with the form determined for 
the teeth on the plane F G (Fig. 87 ), the form obtained on the 
front elevation having O" as center, is elongated. One should, 
therefore, take account, in practice, of the elongated shape of the 
teeth in these wheels compared with those of corresponding plane 
gearings. 

The drawing of the front elevation of the wheel is made in 
exactly the same manner. 

Defects which Present Themselves in these Gearings. 

320. When, in a gearing, the normal to the point of contact 
does not pass through the point of tangency of the primitive cir¬ 
cumferences, the transmission of the force is irregular. 

The faults of construction which most often produce this 
effect, are : 

First—A relative disproportion between the total diameters of 
the two wheels. 

Second—A gearing too close or too slack. 

Third—A bad teeth range. 


Gearings. 


223 


According to the case, one will then find in the gearing a 
“butting” ora “drop.” 

321 . The butting , also called binding , is the irregular contact 
of two teeth before the line of centers. If, for example, a is the 
point of tangency of two 
primitive circumferences O C 

and O' (Fig. 89) and c the 
point of contact of a tooth 
and a leaf, one will find on 
drawing the normal to this 
point that in place of pass¬ 
ing through a it will cut the 
line of centers at a point a' 
situated between a and O'. 

There will result a diminu¬ 
tion of force transmitted at 
this instant for the two fol¬ 
lowing reasons : 

First—In place of a * 

force F' = F ^ one will 

U a ’ 


have only F' = F - ~ g ? -, as i 

much different from the first I 

as the point a! is found f 

nearer to the center O'. (R 

Second—Increase of Fig g9 

the re-entering friction. 


The causes which can produce this defect are generally : 

(1) Two slack a gearing ; 

(2) A pinion proportionally too large ; 

(3) A bad teeth range. 

Fig. 89* shows the case of too large a pinion ; the pitch of 
the gearing is longer than that of the wheel. The tooth B has 
ceased to conduct the leaf and the tooth A enters too soon into 
contact with the succeeding one. As we have said, the moment 
of the force transmitted is, therefore, diminished. 

Fig. 90 shows the case of too slack a gearing. In place of 
entering into contact with the straight flank of the leaf, the tooth 

*Figs. 89, 90, 91, 93 and 94 have the defects that they should represent generally exagge¬ 
rated, in order to make them better understood. One sees also that by the use of a semi¬ 
circular excess for the leaf, such defects are often rendered more appreciable for pinions of 
low numbers ; these excesses should be of semi-elliptical form. 






224 


Lessons in Horology. 


conducts, first, the excess, the normal cuts the line of centers be¬ 
tween the point a and the center O' of the pinion. 

Fig. 91 shows the case of a bad teeth range; the tooth, too 

short, for example, has 


o’ 



its contact with the 
leaf, as in the preced- 
N ing case : the normal 
passes between the 
point a and the center 
O' and one has a dimi¬ 
nution of the force 
transmitted. 

When the above 
defects are not too 
much accentuated, it is 
possible to remedy 
them, in order to ob¬ 
tain a passable gear¬ 
ing ; but, at least in 
the first case, it is im¬ 
possible to arrive at 
absolute perfection. 

If the pinion is 
slightly too large, one 
can touch up the wheel 
in such a manner as to free the teeth range at the base a b (Fig. 92) 
and make it less pointed, after the manner of the English teeth 
range. 

If the gearing is too slack, one increases the diameter of the 
wheel by careful forging. 

If the teeth range is defective, one can try to rectify it by means 
of a suitable ordinary cutter, or, still better, with an Ingold cutter. 

322 . If the first contact of the tooth with the leaf com¬ 
mences after the line of centers, it may happen that at a certain 
moment of the movement the angular speed of the wheel becomes 
proportionately greater than that of the pinion which it conducts. 
This defect is the drop ; it is produced by 

(1) Too close a gearing. 

(2) A pinion proportionally too small. 

(3) A bad teeth range. 


I 

0 

Fig. 90 





Gearings. 


225 


Fig. 93 shows too small a pinion ; the pitch of the gearing of 
the wheel is greater than that of the pinion. When the tooth B 
should cease the contact on the generating circle, the tooth A is 
still found removed from the leaf that it should conduct. The 
tooth B will slip along the flank of the leaf and at this instant the 
normal to the point of contact will not pass through the point 
of contact of the primitive circumferences, but will cross the line of 
centers at a point nearer 
the center of the wheel 
One will, therefore, 
have, in this case, an 
increase of the force 
transmitted. For a 
uniform movement of 
the pinion, the wheel 
will take an accele¬ 
rated movement ; this 
is, technically speak¬ 
ing, a “ drop.” 

Fig. 94 represents 
too deep a gearing, the 
tooth B conducts its 
leaf farther than the 
generating circumfer¬ 
ence ; there is, there¬ 
fore, produced a slip- 

ping of the point of Flg U1 

the tooth against the 

flank of the leaf, the accelerated movement which the wheel 
takes terminates by a drop of the tooth which follows on the 
leaf which it will conduct. The direction of the normal at the 
point of contact shows that v one has, in this case, also an 
increase of the force transmitted. 

Fig. 95 represents the case of a bad teeth 
range of the wheel. The teeth, which are too 
long, drive the pinion leaves farther than they 
should geometrically ; one can thus recognize 
the drop which will be produced. 

A gearing presenting the above defects can 
be corrected by diminishing the height of the 









226 


Lessons hi Horology. 


ogive in such a manner that the teeth drive the leaves a less dis- 
tance or, otherwise, by forming the teeth in such a manner as to 
give them a greater breadth on the primitive circumference. 

323 . On proving, as we have just done, that the gearing 
of a wheel in too large a pinion produces a butting, that, on 


the other hand, too deep a gear¬ 
ing produces a drop, one sees 
that it is best to make a deep 
gearing when the pinion is too 
large. 


O’ 



Reciprocally, a gearing whose 
pinion is too small should be 
relatively shallow. 


324 . A defect which one 
encounters often enough in gear¬ 
ings is that which is occasioned 
by pinions whose leaves are not 
long enough, that is to say, 
pinions which are not cut deep 
enough. If the teeth of the 
wheel are correct, one finds very 
often the point of the tooth in 
contact with the bottom of the 
leaves (core of the pinion). If 



one cannot change the pinion, which is the only means to obtain a 
perfect gearing, the ogives of the teeth must then be shortened, 
either by cutting off the points or by modifying the shape. One 
understands that in these cases absolute perfection exists no longer, 
especially if the number of pinion leaves is small ; since then the 
contact should commence before the line of centers. 

325 . One encounters very often, also, pinions of ordinary 
quality in which the flanks of the leaves are not directed toward 
the center, but are diverted more or less from it. Such pinions 
should be rejected as much as possible if one wishes to preserve 
in the gearing the quality of a flank gearing ; if not, the tooth of 
the wheel would have to be formed by means of a curve described 
as we have indicated (215). 

In a gearing, defective either on account of the shape or direc¬ 
tion of the pinion leaves or the wheel teeth, if one modifies one of 
the two profiles it might happen that one arrives at a correct 






Gearmgs. 


227 


gearing fulfilling all the con¬ 
ditions of a uniform trans¬ 
mission of the force, even 
when the essential character¬ 
istics of the flank or epicy- 
cloidal gearing no longer exist. 
In this case, the entire theory 
of the determination of the 
forms of contact is there in 
order to make us understand 
that one has luckily been able 
to find a combination of forms 
fulfilling the condition estab¬ 
lished, that the normal to the 
successive points of contact 
passes constantly through the 
point of tangency of the primi¬ 
tive circumferences. We know 
that this condition suffices for 
the gearing to be perfect, 
whatever may be the shape of 



I 



Fig. 94 


the profiles established. 


Passive Resistances in Gearings. 

326. General Ideas. We have already indicated that the pas¬ 
sive resistances are forces which naturally present themselves in 

all machines in motion (46). 

These resistances are of 
diverse natures : some pro¬ 
ceed from the bodies them¬ 
selves, from their weight, 
their form, their dimen¬ 
sions, and also from the 
relativeness of the move¬ 
ments which animate them. 
Such are friction, and its 
congenerics, inertia and 
shocks. Others arise, more 
properly, from the medium 
in which these bodies are 
moved, such as, especially, 









228 


Lessons in Horology. 


the resistance of the air. Among these, the principal cause of the 
absorption of work which is to be considered here is the friction , 
of which we will first take up the general study before applying 
the laws to the particular case of the gearings. 

Friction. 

327 . When a body is moved by slipping on another body, 
there is produced a resistance which is opposed to the movement. 
This resistance is due to the action of the two surfaces in contact, 
when the movement already communicated to the body allows the 
inertia to be excluded. This resisting force is friction ; it appears 
to proceed from the reciprocal action of the molecules of the two 
bodies. The inequalities of surface more or less evident in these 
bodies penetrate each other reciprocally, fit into each other with 
much greater intensity in proportion as the two bodies are more 
closely pressed together. Moreover, when one of these bodies is 
displaced the resistance produced by this ‘ ‘ binding ’ ’ is further 
increased by the driving back of the molecules situated in front of 
the moving body. 

328 . Besides this cause of resistance, there exists a second one, 
due to the adhesio 7 i of the two surfaces. The effect produced by 
this second cause can be made very apparent by placing on each 
other two planes of the same kind ; if the surfaces are very care¬ 
fully planed and perfectly polished, as, for example, those of two 
mirrors, the adhesion may become so great that the separation of 
the two bodies becomes very difficult.* 

Friction depends, therefore, on the two causes mentioned ; but 
the last is very often neglected if the two surfaces are directly in 
contact, that is to say, if there is no coating or lubricating sub¬ 
stance, such as oil, between these surfaces. Numerous experi¬ 
ments have, in fact, proven that this resistance may be neglected 
when the extent of the surfaces in contact is not very great. 

329 . But when one interposes a greasy substance between 
the two bodies, it is no longer possible to neglect this last cause, 
which, in certain cases, may diminish the friction properly 
speaking and, in others, increase it. We will treat, further on, 
of this question and will limit ourselves for the moment to the 
study of “ dry friction.” 

* This phenomenon arises from the more or less complete expulsion of the air bet ween 
the surfaces and from the pressure of the exterior atmosphere. 



Gearings. 


229 


330 . The Two Kinds of Friction. If the same part of the 
surface of one of the rubbing bodies always remains in contact 
with the other body, there is sliding, and the friction takes the 
name of “sliding friction.” If, on the contrary, the surfaces in 
contact change at each instant, there is rolling and the friction 
takes the name of “rolling friction.” An example of the first 
case is the friction which is established during the movement of a 
sleigh along a road ; and of the second case, that which is pro¬ 
duced when a wheel rolls on a plane. 

We will occupy ourselves especially with the sliding friction, 
the only kind which we will encounter in horology. 

331. The slipping may be linear , that is to say, be effected 
along a plane or any surface whatever when one of the bodies is 
continually displaced with relation to the other ; or it may be 
circular , if one of these bodies turns on itself without going 
forward, for example, a trunnion in its bearings. 

The friction of the teeth of a gearing is produced by a linear 
slipping ; that of the pivots of these same wheels in the interior of 
the holes in which they turn is produced by a circular slipping. 

332. Laws Of Friction. It has been discovered by very care¬ 
ful experiments that the resistance due to friction is subject to three 
principal laws which can guide in the applications and which are 
sufficiently exact within the limits between which they are con¬ 
sidered in machines. 

First— The friction is proportional to the normal pressure ; 
that is to say, the resistance is always the same fraction of the pres¬ 
sure which applies one body on another, which is easily understood, 
since the actions of the molecules should arise by reason of this 
pressure. 

Second— The friction is independent of the surfaces in contact; 
this is to say, when this extent increases without the pressure 
changing, the total resistance remains the same, although the pres¬ 
sure on each element of surface is found to be diminished in inverse 
relation to the extent of these surfaces. Since, for given sub¬ 
stances, the friction is a constant fraction of the pressure, it fol¬ 
lows that a heavy body drawn on a plane gives rise always to 
the same resistance, whatever may be the extent of the surface 
of contact. 

Third— The friction is independent of the speed of the move¬ 
ment ; which is to say, that the same amount of work is necessary 


230 


Lessons in Horology. 


in overcoming the friction of a body traversing a certain distance, 
no matter what may be the speed which animates the body. 

By the aid of these three fundamental laws and of the values 
determined, experimentally, in order to establish the relation of the 
friction to the pressure according to the nature of the surfaces in 
contact, one may value in each case the work absorbed by friction. 

333. Experimental Determination of the Force of Friction. Let 
us suppose that a body with weight P be acted upon by a force F 
which makes it slide with a uniform movement on a surface A B 
(Fig. 96). One knows that when a body is moved uniformly, 



the algebraic sum of the forces which act on this body is equal to 
zero. The force F should, therefore, be equal, and in contrary 
sense to, the force of friction ; it will be, therefore, the measure of 
the greatness of this resistance. 

One of the laws which we have cited, showing that the friction 
is proportional to the pressure, it follows that if the weight P is 
doubled, the friction is doubled at the same time and consequently 
its equivalent F 

The relation is, therefore, constant for the same substances 
in contact ; this is called the “coefficient of friction,” that is gen¬ 
erally represented in the calculations by the letter /. Thus one has 



When this coefficient is known, as well as the pressure P 
extended normally to the surfaces in contact, one can determine the 
friction F by multiplying the pressure P by the coefficient f. 
Therefore, p _f p 

334 . Let us note that the coefficient of friction does not always 
keep the same value for different surfaces of the same kind, for the 












Gearings. 


231 


harder a body is and the more it is polished, the less is the friction. 
Its value is also modified by interposing a greasy substance, oil, 
for example, between the surfaces in contact. The object of this 
operation is principally to avoid the grating and the heating of the 
frictioning bodies. One knows, in fact, that without this precaution 
there are detached from the surfaces small fragments which groove 
them deeper and deeper ; the friction speedily increases and the 
heating which results from it can even go so far as to make the 
bodies red hot and to set them on fire if they are combustible. 

One finds that friction of steel on steel produces by the grating 
a reddish dust, which is, probably, oxide of iron ; the dry friction 
of steel on brass enables us to prove that a certain quantity of brass 
is deposited on the surface of the steel ; the heating should, in this 
case, be considerable. 

Horologists know the grooves, often very deep, which the lack 
of oil on the pivots produces, when these turn a long time, dry in 
their holes (the fact is especially noticeable on the pivots of the center 
wheel) ; they are familiar also with the deep lines worn in the 
leaves of tempered steel pinions, caused by the teeth of the wheels 
made of gold alloys, which, for this reason, are almost entirely 
abandoned in these days. One sees by these examples that a high 
speed of the mobiles is not necessary to produce the grating, 
which is on the whole entirely in conformity with the third law of 
friction. 

335 . The following table is intended to give an idea of the 
mean value of the coefficient f in the most general conditions. It 
is best, in each particular case, to choose this value properly, 
according to the probable conditions of the action of the parts in 
motion. 

COEFFICIENTS OF FRICTION 


BODIES IN CONTACT 

RELATION / OF THE FRIC¬ 
TION TO THE PRESSURE 

Metal on metal. 

0.15 to 0.17 

Metal on precious stones .... 

O.I5 

Wood on wood. 

0-33 

Bricks and stones on the same . . 

O .65 

Leather bands on metallic pulleys . 

0.30 to 0.40 









232 


Lessons in I loro logy. 


In large machines whose frictioning parts are carefully greased 
the coefficient of friction diminishes to a value of f = 0.08. 

336. Work of Friction. The mechanical work of a force being 
the product of this force by the path traversed by its point of 
application, when the path and the force have the same direction, 
one will have, if E is the path traversed, 


E 


W. F — f P E. 

If the two bodies are movers, it will be necessary to consider 
the two forces of friction which have f P for common value and 
which act on each body in the inverse direction of its movement 
with regard to the other, each of the forces producing work. 

Suppose that the movements of the two bodies A and B 
(Fig. 97) are effected in the direction of the arrows (1), E and E' 

being the respec- 

" > ^ tive paths traversed 

in this direction, 
which will be that 
of the relative 
f P movement of the 

body A, if E is 
greater than E'. 

If we consider 
the movement of 
the body A the friction will produce a work f P. E. } which will 
be negative, since the direction f P is the inverse of that of E ; this 

is, therefore, a resisting work. On the contrary, the work of B on 

A will produce on this first body a positive work f P. E'. f which 
will take away from the resisting work f P. E., so that finally the 
resisting work produced by the friction will be 



IV F = / P E — f P. E' = f P {E — E') } 

E — E’ being positive. If E — E' becomes negative, the equa¬ 
tion no longer holds, and the direction of the friction must be 
changed ; in place of having f P (E — E '), one will have 
fP{E’-E). 

Let us here note that the work developed by the friction on 
one of the two bodies is positive ; with regard to this body, there¬ 
fore, friction plays the part of motive force. This property is 
employed industrially in the transmission of movement by cylin¬ 
ders, cones or friction plates. 









Gearings. 


233 


337. Angle Of Friction. Suppose a body resting on an inclined 
plane A C (Fig. 98) ; let us admit that we have regulated the 
inclination of this plane in such a manner that the body may be on 
the point of moving, or, what amounts to the same thing, that it is 
moved with a uniform motion, the length of this plane. In this 
case the force of the friction is equal and in a contrary direction to 
the force which acts to make it descend. 



The weight p of the body acts along a vertical line c a , pass¬ 
ing through its center of gravity ; making c a — p, drawing the 
line c b parallel to the plane and c d perpendicular to this direction 
one will be able to form the parallelogram of the focus by drawing 
the lines a d parallel to b c and a b parallel to c d. The length c b 
will then represent the value of the force F tending to make the 
body descend along the plane, and the length, c d> the normal 
pressure P. We will, therefore, have 

b c F 
b a ~ P ' 

The similar triangles b c a and B C A give, moreover, the 
proportion b c _ B C _ F 

b a ~~ BA ~ P ■'* 

One can thus see that the coefficient of friction is equal to the 
quotient of the height B C of the inclined plane divided by the 
length of the base A B. 

338 . Designating by <}>, the angle, CAB , that the inclined 
plane forms with the horizon when the movement takes place, the 
two components of the weight p can also be represented by 

p. sin cf> = F 

following the direction of the inclined plane and by 

p cos. cf> = P 









234 


Lessons in Horology . 


perpendicularly to this plane. One has, therefore, 


F 

P 


p sin <J> 


p cos 


= tang <J> = /. 


One discovers, on varying arbitrarily the extent of the surface 
in contact and the weights of the bodies, that the angle of inclina¬ 
tion does not vary for the same substances in contact. 

This angle is called the angle of frictio 7 i and the numerical 
value of the relation of the friction to the pressure, equal to tang. 4, 
is the coefficient of the friction. For hard and polished metals and 
the stones used in horology, this angle has a value varying from 
7° to 8° 30'. 

Example of Application. —On a plate of tempered and 
polished steel we place a ruby, a lever pallet, for example. We 
elevate little by little one of the extremities of the plate until the 
ruby commences to slide with a uniform movement. The height 
B C to which it was necessary to elevate the steel plate being 
13.4 mm. and the length of the base AC 89 mm., the coefficient 
of friction of the ruby on the steel should be 


/ 


i3-4 

89 


0.15. 


The angle of friction will be, in this case, 

tang. 4 = 0.15 
and <{, = 8° 32', 

value that we will adopt in our calculations. 


Calculation of the Friction in Gearings. 

339 . Knowing the normal pressure between the teeth of a 
gearing, one has the value of the sliding friction (333), so that if one 
knows the length of the space traversed by the friction of one tooth 
on another, one would have the work absorbed by this friction (336). 

Before entering into the details of this calculation, let us recall 
the kinetic question of the transmission of the movement. 

We have found that by means of gearings, the movement of one 
wheel is uniformly transmitted to another ; this geometrical demon¬ 
stration is independent of the material of which the wheels are formed, 
of the nature of the friction, etc. This property holds good whatever 
may be the friction in play and the greatness of the efforts which are 
shown. The passive resistances have, therefore, no effect on the trans¬ 
mission of the movement, properly speaking ; they only increase 





Gearings. 235 

the work to be expended in order to produce the movement of the 
motive wheel. 

One thus understands that the work of friction may be generally 
expressed as function of the resisting useful work to which it is added. 
340 . Let us adopt the following notation : 

A, the wheel which controls the movement ; 
r, its primitive radius ; 

7 i } its number of teeth ; 

A', the wheel controlled ; 
r', its primitive radius ; 
n', its number of teeth. 

a = t c = t c’ (Fig. 99) the pitch of the gearing, and Q the resis¬ 
tance opposed to the movement of A acting tangently to the primi¬ 
tive circumference. 



Generally, in gearings, there are several teeth of the wheel 
which work at the same time ; but, in order to facilitate the calcu¬ 
lation, we will suppose that there is only one and that it controls 
from t to c, that is to say, a space equal to the pitch. During this 
passage a , the work absorbed by the normal pressure between the 
teeth is equal to this pressure multiplied by the length of the curve 
traversed by the point of contact t in its passage from t to a 
length which does not perceptibly differ from a*. But, since the 

* Let us note that the normal pressure has not generally a constant value ; it would have 
it only in the case of gearings by involvent of circle, if one neglected the friction. For the 
others, it is variable and it is the mean value of this quantity that must be made to enter into- 
the expression of the work of the friction. 










236 


Lesso?is in Horology. 


point of contact is very slightly removed from the arc t c' , one can 
suppose, without committing an appreciable error, that the pressure 
is equal to the force Q acting tangently to t c\ and the work 
absorbed is, therefore, a q 

From the hypothesis that the normal pressure between the 
teeth is constantly equal to 0 (mean value of this pressure), it 
follows that the friction is equal to 

IQ • 

As to the work absorbed by this friction, one remarks that 
while the point of contact t passes to the space traversed by 
this point on the tooth which controls is equal to t’ c and that 
which it has traversed on the controlled wheel is t' c', from whence 
it follows that there has been sliding on a length equal to 

t' c — t' c\ 

a difference which one can suppose equal to a straight line joining 
c to c'. The work absorbed by the friction is, therefore, 

QfX c c'. 

Dropping the perpendiculars c e and c’ e’ to the line of centers, 
O O’ being almost parallel to c c', one can suppose 

cc' = ee / = fe + te / i 

but, one has 

. F7* , . , A~~r .* 

t e = - and t e' — -7-; 

2 r 2 r 

and since one can admit c ^ _ c r t — a 

which comes nearer the truth as the pitch becomes smaller with 
relation to the radius, one has, therefore, 

t e = —— and t e' - a , ; 

2 r 2 r' 

consequently the work absorbed by the friction is 



The work which the wheel A should transmit to the wheel A' 
for the distance traversed a is, therefore, 

IV„ = Qa + Qate (- 1 - + yy (1) 

* A chord is tho mean proportional bet ween the diameter and its projection on this 
diameter. 










Gearings. 


237 


If P is the motive force which acts tangently on the driving 
wheel, the motive work for the distance traversed a is P a and one has 

Pa= Qa + Q (g- + 

from whence 

p = q q ~ (~+^y 

341 . Taking up again the formula (1) in which Q a represents 
the useful work ‘ ‘ W u , ’ ’ this can still be put under the form 

Wm = Wu + Wu - Y- (2) 

Since one has, moreover, 

r n 

r' n' ’ 

n and n' being the number of teeth, and since 

n a = 2 tt r and n' a = 2 ir r', 

one obtains „ a n f a 

r = -and r' -- 

2 IT 2 IT 

On replacing these values in the formula (2) and simplifying, 
it becomes 

Wm = Wu + Wu - fir (— + (3) 

\ n n' J 

or 

w m = Wu [. + 

from whence one draws 

w _ Wm 

'+/-(■£ + £) W 

342 . On examining these two last equations, we can see that 
one diminishes the friction by increasing the number of teeth. 

Thus, for a wheel of 64 teeth gearing in a pinion of 8 leaves, 
one would have, on admitting W m = 1 and f= 15, 


Wu ■= — 

1 

while for a wheel of 
would have only 


i _ 1 

/ 1 1 \ — 1.066 ’ 

+ 0.15.3-1416 (- 6T + - j -) 

96 teeth gearing in a pinion of 12 leaves, one 


Wu 


1 + 0.15. 




1 

1.044 ' 














23S 


Lessons in Horology. 


One thus discovers the practical rule that the« number of teeth 
must be increased as much as possible in order to diminish the work 
of friction, to have less wear and a smoother motion. 

343 . For interior gear mgs, the formulas (3) and (4) become 

Wm = Wu F W f TT f ”7“*) (5) 

\ n n' / 

and 



One sees that in these gearings, with the same number of teeth 
the friction is less than in the exterior gearings. 

344 . The friction in the rack can also be deduced from the 
preceding formulas, on remarking that the radius of one of the 
primitive circumferences becoming infinite the general expression of 
the friction, for n — 00, becomes 

Wm = Wu + Wu /ir ( 7 ) 

and n 

Wu =-^-(8) 

1 + / ¥ 7 


345 . For conical gearings , on preserving the same relations 

as in the preceding cases, one would arrive at the following 

results, a being the angle formed between the axes of the two 

wheels : ,- 

f a \ 1 , 1 , 2 cos a (9) 


and 


ll r m — Wu -f- IVu 


Wu = 


V £ 


+ 




+ 


2 cos a 
r r' 


Wr 


m 


I +/ 


TT 


Vi 


r + 


12 cos a 
n ' 2 n n' 


(10) 


These gearings are smoother than cylindrical gearings of the 
same number of teeth. 

346 . Friction Before and After the Line of Centers. In the 

formulas which we have established, the influence of the friction is the 
same before as after the line of centers ; this result does not agree 
with those of experience, which show, on the contrary, that the 
friction before the line of centers is more hurtful than that which is 
exerted after the passage of this line. However, since we have 
supposed the pitch as being very small, our results can be con¬ 
sidered, in this case, as sufficiently exact. 

It is certain that it would no longer be the same if the contact 
commenced at a relatively great distance from the line of centers. 














Gearings. 


239 


34(T. Let us examine, for example, the case of a wheel with 60 
teeth gearing in a pinion of 6 leaves, since we know that in horology 
this gearing is one of those in which the contact of the tooth and 



Fig. 100 


the leaf should commence the most in advance of the line of centers ; 
and let us examine successively the four following cases : 

First—The wheel drives the pinion after the line of centers ; 
Second—The pinion drives the wheel before the line of centers ; 
Third—The wheel drives the pinion before the line of centers ; 
Fourth—The pinion drives the wheel after the line of centers. 
348. The Wheel Drives the Pinion After the Line of Centers. 
Suppose O and O' the centers of two mobiles (Fig. 100), P the 





240 


Lessons in Horology. 


moment of the force with relation to the axis of the wheel which 
drives, and Q the moment of the force with relation to the axis of 
the wheel which is driven. 

The wheel 0 is in equilibrium under the action of the force P } 
whose moment is P t of the normal force N' = N whose moment 
is — N. Ob and of the force of friction f N> directed perpendicularly 
to the normal force, and whose moment is — / N O d. One has, 
therefore, P — N O b —f N O d = O. 

On the other hand, the wheel O' is in equilibrium under the 
action of the force Q, whose moment is Q and of the normal reac¬ 
tion N' — A 7- whose moment is — N'. O' b'. The moment of the 
force of friction is null, since its lever arm is equal to zero. One 
has, therefore, q _ N. O' b' — O. 

On dividing the first of these equations by the second and 

simplifying-, one obtains 

P _ Ob+/Od 
Q ~ O' b' 

But, on designating by p the angle b t O — b't O' formed by 
the normal and the line of centers, one will also have 

P r. sin P + f ( r + P ) cos P 


Q 

and, on dividing by sin p, 


sin p 


or also 


P 

Q ~ 

P P 
Qr 


r 

P 


( 


1 + / 


r + P 


cotang 


p > 


= I +/ ( I + cotang p. 


(1) 


Let us remark that if, in this equation, we make the friction 
equal to zero, we will have 


or 


P P 
Qr 

P 


— 1 


Q 


an analogous formula to that which we have established (196, equa. 17). 

If, in the above formula (1), one places Q r = i, one will 
obtain for P P a value superior to unity. 

The angle a (Fig. 100) which the leaf is diverted from the line 
of centers, is the complement of p ; one can, therefore, also write 
the equation (1) : 

P P 


Qr 


— 1 


f ( 1 + it) tang °- 












2 4 I 


Gearings. 


Numerical Calculation.—L et Qr = i,/=o. 15, 

6 


one has 


n ' 

n 


60 


a = 42 0 15' 47", 


165 


' ( 1 + v) = 015 ( 1 + -Ir) = °- 

Log: 0.165 = o.2174839 — 1 

Log : tang a = 9.9584454 

°- 1 759 2 93 — 1 
Number . . . = 0.14994 

We will therefore have the relation 

P P 

-Qj = I-I4994- 

On subtracting the friction, and admitting the moment P = 
1 gr., we would have, in this case, 

P _ 60 

Q ~ 6 ’ 


from whence 


£ 


I * 65 - = al gr ‘ 


On introducing the force of friction, one will have 

P i -1 4 994 X 6 0 

from whence Q 6 

1 




= 0.08696 gr. 


11.4994 

349. The Pinion Drives the Wheel Before the Line of Centers. 

In this case the moment Q becomes the moving power and the 
value Q r will become superior to P r 1 . The formula (1) then 

(2) 


becomes 


Qr 

PP 


- / ( 


1 + 


n ' 

n 


tang 


a ) 


remarking that the sign of the friction is changed. 

Numerical Calculation. —The same data as in the pre¬ 
ceding case, except that we take here P r — 1. We have 


then 


/ (1 + tang, a = 0.14994, 

— f(i+ tan g- a = 0.85006, 


Qr 

P r' 


0.85006 


1.1764. 












242 


Lessons in Horology. 


On subtracting the friction, one would have, if Q = 1 gr., 

Q _ 6 

from whence 


P 


60 


60 


P = Q -g- = 10 gr. 

On introducing the force of friction, one will have 


from whence 


Q 

P 
P = 


1.1764 


60 

= 8.5 gr. 


0.11764 

350. The Wheel Drives the Pinion Before the Line of Centers. 

We have in this case (Fig. 101), reasoning the same as in the 
preceding cases, 


from whence 


or, again, 


p — N. o b = a 
Q — N'. O’ b' + / N'. O’ d = a 

P _ r. sin P 

Q P. sin. P — f ( r -j- r ') cos p ’ 

PP 1 

Qr ~ , 


/ ( 1 + cotang p 


But the angle ^ O O', complement of p, is equal to 

n' 

a — 

n 

since the angles traversed in the same time by the two mobiles of a 
gearing are inversely proportional to the numbers of teeth (176). 
One will, therefore, have, 

P P _ _1_ 

Q r . / n \ / nf \ ( 3 ) 

I - / ( I + -^) tang ( a ^) 

Numerical Calculation. —Let a = 17 0 44' 13", n = 60, 
n ’ =■ 6. 

One has 

a ^ = O.I X 17 ° 44 ' 13 " = 1 ° 46 ' 25.3" 

/(“ 4 - 1 ) = 0.15 X 11 = 1 65 
log. 1.65 = 0.2174839 
log. tang. ( a ~ ) = 8.4908948 

0.7083787 — 2 
Number = 0.051095 









Gearings. 


243 


1 -/(. 


+ £) tan g- (° ii) = 1 — 0.051095 = 0.948905 


1 

0.948905 


= 1.05384 = 


Pr' 
Q r 


If the moment of force Z 5 = 1 gr., we have, without the 


friction, 


^ n n b 

Q — P — = 1 '~z~ — °- 1 S r - 

n 60 



On introducing the friction, one has 

P ri = 1 

® ~ 1.05384 ' n 1.05384 

Q = 0.09489 gr. 


1 

10 


from whence 
















244 


Lessons in Horology. 


351. The Pinion Drives the Wheel After the line of Centers. 

The moment Q becomes the moving power and the formula (3) 


becomes 


Q r 

P r* 


= 1 +/ ( 1 + if) tang - (“tt) 


(4) 


on changing the sign of the friction. 

Numerical Calculation. —One has, from the preceding 
calculation, q r 

— 1 + 0.051095 = 1.051095. 

Without the friction, we will have 


from whence 
and if Q = i gr., 


P = 


Q_ 

p 

P= Q 

60 


tv 

-1 

n 

n 

n' 


= 10 gr. 


With the friction, one will have 

p 6 ° 1 _ 10 

6 1.051095 ~ 1.051095 

and P == 9.514 gr. 

352. Recapitulation of the Preceding Calculations. The 

moment of the motive force acting on the wheel being equal to 
1 gramme, the moment of the resisting force with relation to the 
axis of the pinion should be at the instant of the first contact 
before the line of centers, 

Q = 0.09489 gr.; 

and at the instant of the last contact after the line of centers, 

Q = 0.08696 gr. 

When the pinion drives the wheel, we have found at the 
instant of the first contact before the line of centers, 

P = 8.5 gr., 

and at the instant of the last contact after the line of centers, 

P = 9-514 gr. 

One sees that, in the most usual case, when the wheel drives 
the pinion, the force absorbed by the friction before the line of 
centers differs very little from that which is absorbed after the 
passage of this line, which confirms what we have admitted (261).* 


* In order to be able to compare them in an absolute manner, the above figure should be 
calculated for angles of approach and retreat equal to each other. Which explains why, when 
the wheel drives the pinion, the moment of the force absorbed by the friction before the line 
of centers is inferior to that which we have obtained for the instant of the last contact. If 
one calculated, for the above case, the moment of the force Q for an angle of retreat, 17° 44' 13", 
one would find q _ 0.09498 gr., 

a figure very little greater than that which we have obtained for the same angle before the 
line of centers. 












Gearings. 


245 


One sees also that the smaller the driving wheel becomes with 
relation to the one which is driven, the more also increases the dif¬ 
ference of the resistance before and after the line of centers ; the 
numbers of teeth of the two mobiles should, therefore, be increased 
as much as possible. 

Calculations of the Friction of Pivots. 

353. When the watch is placed in a horizontal position, the 
different mobiles of the train rest on the flat “shoulders” of their 



Fig. 103 


lower pivots ; in the vertical position, these same mobiles rest on 
the cylindrical surfaces of the two pivots. The force of friction is 
proportionate to the pressure which the surfaces in contact undergo. 
In the horizontal positions, the pressure proceeds from the weight 
of the mobile and from the lateral force which presses the pivots 
against the sides of the holes; in the vertical positions, these same 
forces are in action, but on account of the position of the gearing 
on the axes, these pressures will generally be different on each pivot. 

354. Work Absorbed by Friction on the Plane Surface of 
the Shoulder Of a Pivot. Suppose (Fig. 102) r” the exterior 
radius of the shoulder and r' the radius of the pivot. Repre¬ 
senting also the mean radius r —J— by 8 and the width of the 
shoulder by /, then l = r" — r. 















246 


Lessons in Horology. 


One will have, consequently, 

r" = 8 -f — and r' — 8 — 

2 2 

The surface of the circular crown with radius r" — S being 

it ( r //a — S' ), 


one will have 

ir ( S n — S' ) 


7 T 


[(• + !)'-(• 


The surface of the circle with radius r' will be 



/ 

/V 

/ /* „ \ 

ir S' = 

= * ( 

8 “ 2 ) =' 1 

( 8! + 4 - 8 0 


P being the pressure exerted on the crown and admitting that this 
pressure varies proportionately to the extent of the surface ; that 
which is exerted on a circle with radius S should be 


P 

2 8 7 




and that which would take place on the circle with radius r" 
would be p 


P + + ~ - 8 / ). 


The work absorbed by the friction of the crown is equal to 
the work absorbed by the friction which would be produced on the 
total surface of the circle with radius S’ = 8 + 2 diminished by 
that which would take place on the surface of the circle with 
radius S = 8 — -5 ; it is, therefore,* 


«'=t*/[7(s+~) +jD 




or 

IV^lTTfP 


S 2 -|-8 / 

8 4- L -1- 4 -. 

^ 2 ‘ 28/ 


1 


[ (8 + i) _ (S _ £)] j 


*For a pressure P, the force developed by friction is / P; the work of this force is the 
product of / P by the distance traversed. This distance is not the same for all the points of 
the surface: null at the center, it attains its maximum at the exterior circumference. To 
obtain its mean value, divide the circle with radius r into a number n of equal sectors 
sufficiently small so that each of them can be regarded as a triangle. The resultant of the 
elementary pressures supported by each of these triangles should pass through its center of 

. p 

gravity, say, at % of the radius. The force of friction being for one of them / ^pthe work 
of this force will be, for one revolution, 

f ~ ~ 2 IT | r=/~ - 4 w r . 

n 6 n 3 

and for the sum of the n sectors, the work will be 


W - f f Prr r. 







Gearings. 


247 


but since 

/ c /\ / 


one has 

( 8 + 2) ~ ( 

8 - 

But 

w= 1 »//>(« + L 

+ 


P 



S' +-8/ 

4 

8 . 


2 8 

2 


it becomes, therefore, 



+ 


r 

8 8 



W=\*fP (§ s + = 2 ,^(t + * 5 ). 


355. In the horizontal position, the pressure P arises from the 
weight of the mobile ; this pressure is always much inferior to the 
lateral pressure with which the pivots are pressed against the sides 
of the holes. Thus, in the preceding equation, one can neglect 
the term 

1 P 

and one has simply 12 8 

IV = f P. 2 ir h = f P (r // -f V). 

356. In the vertical position of the watch, the pressure P on 
the shoulder of the pivot is null ; one can, therefore, also admit 
that the friction is null. 

The work of the friction of the cylindrical surface of the pivots 
against the sides of the holes is expressed by 

W = / P. 2 IT P, 


r' being the radius of the pivot. The formula includes the work 
absorbed by the two pivots, since P is the total pressure and since 
the friction depends only on this pressure and not on the extent of 
the surfaces in contact. 

35 1. Determination of the Lateral Pressure Received by the 
Pivots of the Mobiles in a Train. Let us examine, for example, 
the third wheel of a watch. This mobile receives on one side an 
action on the part of the center wheel gearing in its pinion and, 
on the other, a resistance arising from the pinion of the fourth 
wheel in which the third wheel gears. These two efforts show 
themselves by a pressure on the axis, and the two pivots are pressed 
against the sides of the holes ; for each of these, the load which 
they receive can be represented in magnitude and direction by the 








Lessons in Horology. 


248 


resultant of the partial forces that the axis of the wheel receives at 
each of its extremities. This pressure depends on the relation of 
the distance of the pivots from the point of application of the forces 
in play to the length of the axis. 

358 . Let us imagine the point of contact of the teeth and of 
the leaves on the line of centers and let us represent by P the force 
that the leaf of the pinion receives on the part of the wheel 
tooth of the center wheel ; since, simultaneously with this, one 
wheel tooth of the third wheel is pressed against a pinion leaf of the 
fourth wheel, the force P gives birth to the reaction P’. The 
direction of the forces Pand P' is perpendicular to the line of centers. 

Let us call r and P the primitive radii of the wheel and pinion, 
we will have for the state of equilibrium 

p r , = pr r and p/ = p r L. 

r 


The equilibrium will not be disturbed if one will apply to the 
point diametrically opposed to that of the contact of the tooth and 
leaf a force P 1 equal, and in a contrary direction, to P and, like¬ 
wise, at the point opposite to the point of application of P\ a 
force P' 1 equal and in a contrary direction to this one. 

The resultant of the forces P and P 1 is 2 P; it should be 
applied at the axis of the pinion, parallelly to the components 
(Fig. 103). 

The top pivot of the third wheel will receive on the part of the 
resultant 2 P a force p and the lower pivot a force q in such a way 
that one would have 


One should have 


2 P = p + q- 

P a = q b, 


a representing the distance of the shoulder of the top pivot from 
the middle of the thickness of the center wheel, and b the distance 
from this last point to the shoulder of the lower pivot. 

To determine the values p and q, we extract from the first 
formula q = 2 p - p, 

from whence, on substituting this value in the second, 


and 


p a - 
p = 2 P 


2 P b — p b 
b 


In an analogous manner, one has 

q = 2 P 


a -f b’ 


a 


a -\- b 





Gearings. 


249 


1 



Fig. 103 





















































250 


Lessons in Horology. 


The pivots receive, moreover, an action from the forces P’ and 
P\, whose resultant is 2 P', and we will have 

2 P' =p' + /, 

p' and q' being the strains received by each pivot on the part 
of 2 P f . 

One will thus have # g 

p' = 2 P' — 

and 

' = 2 P' 


a -f- b 
d 

q' = 2 j" — j-j-, 

a -+- 0 

on remarking that d + e — a -j- b. 

Since 


one will also write 

p/— 1 P- 


P' = P 


P 


P 


- - ; —, and q' — 2 P — - -j— 7 - 

a + b v r a + b 

The top bridge receiving, therefore, the two forces p and 
their common resultant will be, in magnitude and in direction, the 
diagonal of the parallelogram constructed with these lines as sides. 

Let us notice that since we have added to P and P' the forces 

■< 

P 1 and P\ equal each to each, the real resultant R should be half 
of that which is thus determined. 

To calculate this value, let us represent by a the angle formed 
by the lines from the centers having their meeting point at O ; we 
will thus have the angle O B A equal also to a. We will have in 
the triangle O B A, O B = p ’ and B A — /, since O A — 2 R ; 
therefore, 


2 R = + \ P* - f y> /a — 2 p P r COS a 


\ 

or, on replacing p and p* by the half of their value, one obtains 


R =yl p (^h -)'+ (r-r - TTi y - ^ ^ 


or 


(1) 


T b 

R = 


+ b 


ci y b 


cos a, 


P 

a + b 




1 P 

2 e b - - cos a. 

r 


For the lower pivot, one would obtain in an analogous manner 
R ' = -^r b aJ a-' + ( ~ ■ d ) - 2 a d SL . cos a. 

359 . Upon examining the Fig. 103, one will notice that the 
pressure of the center wheel teeth is greater than the resistance 
which the leaves of the third pinion oppose. Consequently, for 



















Gearings. 251 

the top pivot of the third wheel, the force O C = p will be greater 
than the force OB— p'. 

For the lower pivot, the pressure of the center wheel diminishes, 
the force O C' = q becomes weaker and O B' = q' stronger. 
There necessarily results a different direction of the resultants R 
and R l} which the construction of the parallelogram of the forces 
shows. 

The direction of these resultants is important and enables us to 
explain the reason why one encounters, in making repairs, pivot 
holes enlarged by wear in a direction often very different from that 
where it would seem that this wear should logically be produced. 
This fact is noticed in clocks or pocket watches whose holes are not 
jeweled. 

360. Let us note, moreover, that the problem we have just 
dealt with is based on the case of flank gearings when the con¬ 
tact takes place on the line of centers ; the normal forces are 
then perpendicular to the radii. But when the contact between 
the tooth and the leaf is displaced or when the gearing is of 
another sort, the normal forces take other directions and the 
value of the resultants, as also their directions, can undergo a 
slight change. 

361. The equations (1) and (2) show that, all other things 
being equal, the pressures become greatest for an angle a = 180°, 
for which cos a = — 1 ; the sign of the last term placed under the 
radical becomes then positive. From this standpoint it would, 
consequently, not be desirable to construct a train, all the mobiles of 
which would be in a straight line. 

362. We establish also that the more the value of r increases, 
the more the pressure diminishes ; this is one of the reasons why it 
is good to increase as much as possible the diameter of the center 
and third wheels, inertia having not yet appreciable influence on 
these mobiles. 

363. Finally, it may not be useless to observe that the relation 
— cannot be replaced by the relation of the numbers of teeth, since 
the mobiles with radii r and F do not gear together, but are 
mounted on the same axis. 

364. Numerical Example. —Suppose 

p = 77.5 gr. a — 0.8. b = 4.06. d — 3.2. e = 2.2. 
r' — 0.87. r = 5.49. a = 95 0 . 


252 


Lessons in Horology. 


We will have for the top pivot 

= 15-95 


and 


P 

a + b 


77-5 

4.86 


0.87 

5-49 


= 0.16, 


b* = 4.06*= 16.40. € ^) = (°‘ l6 X 2.2)* = 0.1239. 


7 

2 a d - - cos a = 2 X 2 - 2 X 4- 06 X 0.16 X — 0.08715 = — 0.25. 

r 

and R = r 5-95 V 7 16.48 + 0.1239 + °- 2 5 

P = 15-95 V 7 16.8539 = 65.47. 

For the lower pivot, one will have successively 


Pi = 15-95 
and 


^/o.8 2 + 0.16 X 3-2 2 — 2 X 0.8 X 0.32 X 0.16 X cos 95 0 

Pi — J 5-95 V 7 °-64 + 0.26 -f- 0.07 


Pi = 15-95 V o-97 


15 - 71 - 


365 . Let us now determine the value of the work of friction of 
the third wheel’s pivots during one oscillation of the balance. We 
have the formula (356). 

IV. F = f P r x p, 


in which P represents the pressure, r 1 the pivot’s radius, p the 
angle traversed during one oscillation. 

Let us first seek this latter angle. The fourth wheel makes 
one rotation in 60 seconds or in 300 oscillations. If the third 
wheel has 75 teeth and the fourth pinion 10 leaves, this pinion 
turns 7.5 times faster than the third wheel ; the third wheel, 
therefore, makes one turn in 300 X 7-5 oscillations — 2250 
oscillations. 

During one oscillation it will traverse, therefore, an angle p : 



o° 9' 36 "; 


this angle expressed in length of arc with radius equal to unity is 
or o° 9' 36" — 0.00279 

0.0028 in round numbers. 


The diameter of the pivots being 0.26, we will have for the 
top pivot 

W. F — 0.15 X 65.47 X 0-13 X 0.0028 = 0.0035 gr. mm., 
and for the lower pivot 

IV F — 0.15 X 15 71 X 0.13 X 0.0028 = 0.00085 gr. mm. 











Gearings. 


253 


The total work absorbed by the friction will, consequently, be 
IV. F — 0.0035 -f 0.00085 = o-oc>435 go mm. 

The work of the motive force applied to this wheel being 
o. 21 gr. mm. during the duration of one oscillation of the balance, 
one can prove that the work absorbed by the friction of the pivots 
represents about the fiftieth part of it. 


Influence of the Oil. 

366. We have said at the beginning of the study of friction 
that the introduction of a greasy substance between the frictioning 
surfaces of two bodies, compelled to slide on each other, is neces¬ 
sary in all cases where a heating and consequently grinding and 
wear are to be feared. 

When greasy substances are interposed between two surfaces, 
these are no longer in immediate contact, the molecules of grease 
form little spheres which roll between the two bodies. In most 
cases, especially in large mechanisms, the friction will be reduced 
by this fact. 

In horology, especially in pocket watches, the inverse phe¬ 
nomenon can present itself. The oil which is used introduces a new 
resistance, an adhesion, or, otherwise expressed, a “sticking.” 
This new resistance is added to the friction and it can happen that 
the coefficient of the sum of the two resistances may be greater than 
the coefficient of dry friction. With regard to the weak forces in 
action on the last mobiles of the train, on those of the escapement 
and on the balance, this last resistance cannot be neglected. Unfor¬ 
tunately, it is very difficult to express this force in figures, because 
it depends on the nature of the lubricant, on its degree of fluidity 
and on its unchangeableness. 

The friction which is exerted through the agency of a lubricant 
depends on the speed of the bodies in contact, and on the extent of 
their surfaces. It depends also on the nature of the movement ; 
thus, it is different on an annular balance when the latter is animated 
with a continuous circular movement and when it is animated with 
an oscillatory movement (circular reciprocating). One under¬ 
stands, in this latter case, that a certain quantity of oil participates 
in the movement of the pivots and that this oil would have a ten¬ 
dency to continue in the direction of the movement, although the 
pivots turn already in the opposite direction. 


254 


Lessons in Horology. 


In all the experiments relative to the friction of lubricated 
bodies, care must be taken to assure oneself that the lubricants are 
neither altered nor expelled. 

One can take as a general principle that the best lubricant is 
that which is the most fluid, that is to say, it is better, when one 
can, to replace grease by oil, oil by water, water by air, which is 
equivalent to suppressing all lubricants. This supposes that the 
speed of the mobiles may be sufficiently great not to expell the 
lubricant experimented with. But a considerable speed is necessary 
for the pieces to retain a fluid lubricant like water and with still 
much more reason for them to leave between themselves a sufficient 
cushion of air. Experiments have been made with the astonishing 
result of showing the friction almost suppressed between two pieces 
rubbing together without any lubricant and at an enormous speed.* 

This almost entire disappearance of friction is due to the inter¬ 
position of a cushion of air, a perfectly elastic matter, between the 
surfaces in contact. 

In horology, in all cases where the use of a lubricant is neces¬ 
sary, one must, therefore, take into account the speed of the 
mobiles and the pressure which they have to support. Thus, the 
wheels of the stem-winding mechanism should always be greased 
by means of a semi-fluid lubricant.f 

The motive spring, as well as the pivots of the arbor around 
which the barrel turns, should be lubricated with a thicker oil than 
that which one employs for the train and the escapement. 

The principal qualities of the refined oil which is used in 
horology should be its unchangeableness by the atmosphere and by 
the various temperatures which the watch must stand, its perfect 
flnidity and the absence of acids in its composition. The solution 
of this question so important to the preserving, for the longest pos¬ 
sible time, of the precision in the running of chronometrical instru¬ 
ments, lies within the domain of organic chemistry. 

*Hirn’s experiments. 

f Practical men value highly for these mobiles, a mixture of pure white wax and refined 
animal oil. 



Gearings. 

Application of the Theory of Gearings. 


255 


Functions of the Heart in Chronographs. 

367 . Chronographs are horary instruments intended to measure 
very small intervals of time. For this purpose these watches are 
furnished with a special hand fastened at the center of the dial and 
traversing a division generally exterior to the minute circle. The 
shortest interval of time measured by the hand of the chronograph 
is equal to the duration of one oscillation of the balance ; thus, 
when the balance of the watch beats 18000 oscillations per hour, the 
chronograph indicates the duration of an observation to about one- 
fifth of a second.* 

These mechanisms are of many different constructions ; their 
movement is controlled by the train of the watch, causing, by this 
fact, a slight additional burden to the motive power. Before the 
observation, the chronograph hand is fastened and remains on the 
division zero. On pressing an exterior push-piece, this hand is 
immediately put into motion ; at the end of the observation, a 
second pressure stops the hand and, finally, after reading, a third 
pressure brings it back suddenly to the division zero, where 
it remains held in place until the moment of a new use of its 
function. 

The invention of the chronograph goes as far back as the year 
1862 and is due to Adolphe Nicole, originally from the valley of 
Lake Joux but established in business in London. 

368 . It does not belong to the plan of this study to give a 
description of the mechanism composing this instrument; it will 
suffice for us to show that the action which returns the hand 
to zero on the division, is effected by the fall of a jumper on 
a heart-shaped eccenti'ic fastened on the axis of the wheel which 
carries the chronograph hand. We will especially occupy our¬ 
selves here with the determination of the form to be given this 
eccentric. 

The condition which the heart should fulfill is to present, at 
every point of its outline, a sufficient inclination to the lever (or 
jumper) which works it to assure the slipping of the lever as far as 
the origin of the curve. 

*C. W. Schmidt, a Swedish engineer, living in Paris, has constructed chronographs 
indicating the duration of a fraction of an oscillation ; these apparatus, with electro-magnetic 
release and stop, are intended especially to measure the speed or projectiles. 




256 


Lessons in Horology. 


On imagining the axis on which the heart is fixed animated 
with a continuous circular movement and the extremity of the 
lever constantly pressed against the exterior border of the curve, 
the problem becomes, to find the form capable of changing, 
uniformly, a continuous circular movement into a reciprocating 
circular movement. 

369 . Let us first examine the simpler case of the transforma¬ 
tion of a continuous circular movement into reciprocating rectilinear 
movement, by means of a heart-shaped eccentric. 



Since the reciprocating movement should be uniform, the 
point B of the line A B (Fig. 104) should successively occupy the 
equidistant positions B, 1, 2, 3, 4, ... . A , the lengths B 7, 12 , 23 , 
.... being supposed equal fractional parts of the total 
course B A. If, from the point O as center, one described cir¬ 
cumferences passing through the points B , /, 2, 3, 4, . . . . A , 
and if one divides the circumference whose radius is O A into 
the same number of divisions into which the line A B has 
been divided, the intersections of the circumferences, with the 
radii passing through the points of division, will indicate succes¬ 
sively the points through which should be described the envelope 
curve of the point B. By construction, the uniform movement 









Gearings, 


257 



Fig. 105 






Lessons in Horology. 


258 

of this curve around the center O will communicate a uniform 
rectilinear movement to the point B, alternately from A to B and 
from B to A , the lower part of the curve naturally being symmet¬ 
rical with the upper part. One recognizes thus that the form 
obtained is that of a spiral of Archimedes , elongated, whose equa¬ 
tion is 

8 = a w -j- C ; 

the radius vector S less a constant quantity, C, is always propor¬ 
tional to the angle described, w. 

3<r0. In chronographs, the reciprocating movement of the 
point B is not rectilinear ; it is circular and its movement is 
executed around a center O' (Fig. 105). 

Let us, therefore, now determine the shape of the heart 
suitable for the above new condition, and let us make the 
pointed end of a lever O' B traverse an arc A B with a uniform 
movement, while the axis O turns the arc B D C = •ir r 0 . Let 
us divide the arcs A B and B D C into the same number of 
equal parts, and let us describe from the center O concentric 
circumferences passing through the division points of the arc A B ; 
draw afterward the radii passing through the division points of the 
arc B D C. 

In order to determine, now, the points n, <?, A' and 
m\ n ’ , o', of the curve of the heart, let us consider, as in the 
preceding case, the intersections of the circumferences and of the 
radii, but let us lay off here in addition the lengths of the 
arcs included on each circumference between the initial radius 
B E and the arc described by the point B of the lever, before 
the points of intersection considered. One would thus obtain 
the curve B m n o A' o’ n’ m’, which would fulfill the con¬ 
ditions desired. In fact, supposing, for example, the point o of 
the heart arrived at the point 3 of the arc B A , the radius 
O K will be superposed on the radius B E, the axis of the 
heart has completed three-quarters of its course and in the same 
time the point B of the lever will have been lifted up three- 
quarters of the distance B A. 

3 H. Let us decide to determine by calculation the value 
of the radius vector of the heart corresponding to the point 
of the lever for any position whatever of the axis of the eccen¬ 
tric (Fig. 106). 


Gearings. 


259 



Fig. 106 














26o 


Lessons in Horology. 


Suppose : 

R , the distance between the center of rotation of the heart and 
that of the arm (R should also be the distance from the point 
of contact to the center of the arm) ; 
r, the variable radius vector from the center of the heart to any 
point whatever of the exterior curve ; 
r Q , the radius vector of the heart corresponding to the position of 
repose, when the hand of the chronograph is at zero ; 
r* t the greatest radius vector of the heart ; 
a, the angle formed by the radii vector r 0 and r ; 

0, the angle formed by the line of centers O O' and the radius R 
of the arm when the latter is in contact with the radius vector r 
of the heart ; 

0 O , this angle when the arm is in contact with the radius r 0 ; 

O', this angle when the arm is in contact with the radius r'. 

The radius r is, therefore, the chord of an arc of a circle with 
radius R and corresponding to angle 0 ; one can write 

r — 2 R. sin \ 0. 

The angle a, formed by the radius vector r Q and the radius 
considered r, is different from the angle which the heart should 
turn starting from the position of repose to the instant when the 
radius r coincides with the point of the lever. Designating this 
last angle by *y, we have in effect 

H = a =t P, 

according as the axis turns to the left or to the right, p being the 
angle formed by the fixed direction of the radius r Q and that which 
the radius r takes at the moment of contact with the lever. Let us 

P lace P = l(«-»o). 

this angle being inscribed in the circumference with radius R 
and 0 — 0 O angle at the center embracing the same arc. We will 
have, consequently, ^ _ Q ^ e _ e o ) 

Let us now establish the relation which connects the angles 
0 — 0 O and *y, angles which should be in a determined relation, by 
virtue of the mechanical principle stating that the transmission of 
the force is uniform when the angles traversed in the same time by 
two mobiles which drive each other remain constantly in the same 
relation. But when the lever traverses the total angle 0 / _ 0 O , 


Gearings. 


261 


the heart executes a demi-turn, therefore, an angle equal to ; one 
will, therefore, have „ _ 

' y TV 

from whence 6 6 ° 0 0 ° 

_ * ( 6 — 9<Q 

y O' — 0 * 


Consequently, one can write 

0 — 0 

a = ir 

and 


0' — 0 


°-=M(0-e o ) 


0 


ir 


and also 


0' — 0, 


TV 


0' - e o ^ T 




1 w o 


0 0 

a “h ST °fl " ~+- ? ®o = ' Ir ST-0 + i ®) 


0 ' — 0 , 


0' — 0, 


from whence 


TV 


O ' — 0 , 


=F \ 0 O 


( r=er ^ 0 *• 


One will thus have 


a 4 - ir 


9 = 


0 ' — 0 


IT 


=F \ 0 O 


0 ' — 0 , 


u ± i (r - 0o) 


+ ®o- 


0' — 0, 


The equation of the two branches of the curve of the heart 
expressed in polar co-ordinates will, therefore, be 


r = 2 R. sin \ 


TV 


+ e< 


_ V - 1 

0 ' — e - * 


372. Numerical Calculation. —Let us admit 


r Q = 2 R. sin \ 0 O = 4. 


2 R. sin \ 0 ' = 24. ./? = 140. 


We will have 4 , . . 0/ 24 

sin \ 0 O = -5- and sin £ 0' = -5-, 

1 • 1 1 0 280 280 

which gives 

0 O = i° 38' 13.6* 
consequently, 0 — 95 ° 3 . 

O' — 0 O = 8° iL 49.4". 

Expressed in seconds of the arc, the angles 0 O , 0' — 0 O and » give 

0 O = 5893.6 seconds 

0 ' - 0 O = 29509.4... “ 

TV — 684OOO ... “ 

Let us first calculate this equation under the form 


r — 


— 2 R sin £ 


"_+ -_0 O ) 
o' — 


V + 0 C 


and suppose a = 30°. 



















262 


Lessons in Horology . 


We will have 


i («'-«.) = 


7T = 648OOO" 

14754.7 


log 


( **■ 4“ Z (®o) ) 

— log : ( 0 ' — 0 ) 


— log: 


662754.7 
The angle a = 
log : 108000 

* + i (0 / - 0 O J 
0' - 0„ 


5.8213528 

4.46 99589 

1.35 1 3939 


= 30° — 108000" 

= 5.0334238 

= 1.3513939 

3.6820299 log: 4808.72". 


4808.72 -f 0 O = 4808.72 -f 5893.6 = 10702.32"; 

therefore, 0 = 2° 58' 22.32" 

|0=i° 24 11.16" 


from whence 


log ■ sin ^ 0 — 8.4139741 
log : 2 R — 2.4471580 

log : 2 R sin \ 0 = 0.8611321 
r = 7.26327 for 0 = 30°. 


Similar calculations will give successively 


r = 10.5254 for a — 6o° 

r — 13.7862 “ a = 90° 

r — 17.0451 “ a = 120° 

r — 20.3017 “a = 150° 

r — 23.5623 “ a = 180 0 


For the other branch of the curve the formula would be 


r 


2 R sin \ 




f 


and identical calculations to the preceding would give the following 
results : 

r = 7.4153 for a = 30° 

r = 10.8294 “ a = 6o° 

r — 14.240 “ a = 90° 

r = 17.66 “ a — 120° 

r — 21.06 “ a = 150° 

r — 24.4648 “ a = 180 0 


For this last calculation, the radius r no longer belongs to the 
closed curve, but rather to the prolongation of this curve. Let us 
remark that the greatest radius vector should be equal to 24. 












Gearings. 


263 


If one would like to know, also, the value of the angle p, cor¬ 
responding to the above data, one would have 


P = Kfl'- e G ) 


/ _ e x _ 8 ° ii 7 4 9 - 4 " _ 


= 4 ° S' S 4 -l"' 


373 . In order to obtain a greater stability for 
the chronograph hand, one prefers, sometimes, to 
make the heart like the adjoining form (Fig. 107); 
one suppresses, in doing this, a part of its curve, 
but, by way of compensation, the part B A B’ of 
the arm is terminated by curves fulfilling the 
conditions desired. 



Fig. 107 


Note. —In the original French, paragraphs 374 to 377, inclu¬ 
sive, appeared at the end of Vol. I as an appendix, but in trans¬ 
lating were moved to their proper places in the text and the 
paragraph numbers, together with the figure numbers under 
cuts, were accordingly changed as follows : 

Paragraph No. 374 now appears as No. 169 a on page 114 
Paragraph No. 375 now appears as No. 169 6 on page 115 
Paragraph No. 376 now appears as No. 169 o. on page 115 
Paragraph No. 377 now appears as No. 226 a on page 154 
Paragraph No. 378 now appears as No. 226 b on page 155 
Figure No. 108 now appears as No. 37 a on page 114 
Figure No. 109 now appears as No. 61 a on page 155 





THE 


WATCH ADJUSTER'S MANUAL 



A Complete and Practical Guide for 
Watchmakers in Adjusting Watches 
and Chronometers for Isochronism, 
Position, Heat and Cold. 


By Charles Edgar Fritts (excelsior), 

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THE ART OF ENGRAVING 



A Complete Treatise on the Engra¬ 
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to Letter and Monogram Engraving. 
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THE OPTICIAN’S MANUAL 

VOL. I. 


By C. H. Brown, M. D. 

Graduate University of Pennsylvania; Professor of Optics and Refraction ; formerly 
Physician in Philadelphia Hospital; Member of Philadelphia County, 
Pennsylvania State and American Medical Societies. 


The Optician’s Manual, Vol. I., has 
proved to be the most popular work on 
practical refraction ever published. The 
knowledge it contains has been more 
effective in building up the optical profes¬ 
sion than any other educational factor. 
A study of it is essential to an intelligent 
appreciation of Vol. II., for it lays the 
foundation structure of all optical knowl¬ 
edge, as the titles of its ten chapters show : 

Chapter I.—Introductory Remarks. 

Chapter II.—The Eye Anatomically. 

Chapter III.—The Eye Optically; or, The Physiology of Vision. 
Chapter IV.—Optics. 

Chapter V.—Lenses. 

Chapter VI.—Numbering of Lenses. 

Chapter VII.—The Use and Value of Glasses. 

Chapter VIII.—Outfit Required. 

Chapter IX. —Method of Examination. 

Chapter X.— Presbyopia. 

The Optician’s Manual, Vol. I., is complete in itself, and 
has been the entire optical education of many successful opti¬ 
cians. For student and teacher it is the best treatise of its kind, 
being simple in style, accurate in statement and comprehensive 
in its treatment of refractive procedure and problems. It merits 
the place of honor beside Vol. II. in every optical library. 

Bound in Cloth—422 pages—colored plates and illustrations. 

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THE OPTICIAN’S MANUAL 

VOL. II. 


By C. H. Brown, M. D. 

Graduate University of Pennsylvania; Professor of Optics and Refraction: formerly 
Physician in Philadelphia Hospital; Member of Philadelphia County, 
Pennsylvania State and American Medical Societies. 



The Optician’s Manual, Vol. II., is 
a direct continuation of The Optician’s 
Manual, Vol. I., being a much more 
advanced and comprehensive treatise. 
It covers in minutest detail the foui 
great subdivisions of practical eye refrac¬ 
tion, viz : 

Myopia. 

Hypermetropia. 

Astigmatism. 

Muscular Anomalies. 


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up to date on these subjects, treated by the master hand of 
an eminent oculist and optical teacher. It is thoroughly prac¬ 
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methods of correction thoroughly elucidated. 

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optics, and the knowledge contained in it marks the standard 
of professionalism. 

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PHYSIOLOGIC OPTICS 

Ocular Dioptrics—Functions of the Retina—Ocular 
Movements and Binocular Vision 

By Dr. M. Jscherning 

Director of the Laboratory of Ophthalmology at the Sorbonne, Paris 


AUTHORIZED TRANSLATION 

By Carl Weiland, M.D. 

Former Chief of Clinic in the Eye Department of the Jefferson College Hospital, 

Philadelphia, Pa. 


This translation, now in its second edition, is the most masterful 
treatise on physiologic optics. Its distinguished author is recognized in 
the world of science as the greatest living authority on this subject, and 
his book embodies not only his own researches, but those of several 
hundred investigators, who, in the past hundred years, made the eye their 
specialty and life study. 

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and his book, as revised and enlarged by himself for the purposes of this 
translation, is the most valuable mine of reliable optical knowledge within 
the reach of ophthalmologists. It contains 380 pages and 212 illustra¬ 
tions, and its reference list comprises the entire galaxy of scientists who 
devoted their researches to this subject. 

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Astigmatism, Aberration and Entopic Phenomena, etc.—in fact, the entire 
book contains so much that is new, practical and necessary that no 
refractionist can afford to be without it. 

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Price, S2.50 (I0s.5d.) 


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THE 

PRINCIPLES OF REFRACTION 

in the Human Eye, Based on the Laws of 

Conjugate Foci 

By Swan M. Burnett, IV?. D., Ph. D. 

Professor of Ophthalmology and Otology in the Georgetown University Medical School; 
Director of the Eye and Ear olinic, Central Dispensary and Emergency 
Hospital ; Ophthalmologist to the Children’s Hospital and to 
Providence Hospital, etc., Washington, D. C. 


In this treatise the student is given a condensed but thor¬ 
ough grounding in the principles of refraction according to a 
method which is both easy and fundamental. The few laws 
governing the conjugate foci lie at the basis of whatever pertains 
to the relations of the object and its image. 

To bring all the phenomena manifest in. the refraction of the 
human eye consecutively under a common explanation by these 
simple laws is, we believe, here undertaken for the first time. 
The comprehension of much which has hitherto seemed difficult 
to the average student has thus been rendered much easier. This 
is especially true of the theory of Skiascopy, which is here eluci¬ 
dated in a manner much more simple and direct than by any 
method hitherto offered. 

The authorship is sufficient assurance of the thoroughness 
of the work. Dr. Burnett is recognized as one of the greatest 
authorities on eye refraction, and this treatise may be described 
as the crystallization of his life-work in this field. 

The text is elucidated by 24 original diagrams, which were 
executed by Chas. F. Prentice, M. E., whose pre-eminence in 
mathematical optics is recognized by all ophthalmologists. 

Bound in Silk Cloth. 

Sent postpaid to any part of the world on receipt of price, 

Si.OO (4s. 2d.) 


Published by THE IvEYSTONE, 

THE ORGAN OF THE JEWEURY AND OPTICAL, TRADES, 
19TH and Brown Sts., Philadelphia, U.S.A. 




SKIASCOPY 

AND THE USE OF THE RETINOSCOPE 



A Treatise on the Shadow Test in 
its Practical Application to the 
Work of Refraction, with an Ex¬ 
planation in Detail of the Optical 
Principles on which the Science 
is Based. 


This new work, the sale of which has already necessitated 
a second edition, far excels all previous treatises on the subject 
in comprehensiveness and practical value to the refractionist. 
It not only explains the test, but expounds fully and explicitly 
the principles underlying it—not only the phenomena revealed 
by the test, but the why and wherefore of such phenomena. 

It contains a full description of skiascopic apparatus, 
including the latest and most approved instruments. 

In depth of research, wealth of illustration and scientific 
completeness this work is unique. 

Bound in cloth; contains 231 pages and 73 illustrations 

and colored plates. 

Sent postpaid to any part of the world on receipt of $1.00 ( 4 s. 2d.) 


Published by The Keystone, 

*HE ORGAN OF THE JEWELRY AND OPTICAL TRADES, 
19TH and Brown Sts., Philadelphia, U.S.A. 









OPHTHALMIC LENSES 

Dioptric Formula for Combined Cylindrical Lenses, 
The Prism-Dioptry and Other Original Papers 

By Charles F. Prentice, M. E. 


A new and revised edition of all the original papers of this noted 
author, combined in one volume. In this revised form, with the addition 
of recent research, these standard papers are of increased value. Com¬ 
bined for the first time in one volume, they are the greatest compilation 
on the subject of lenses extant. 

This book of over 200 pages contains the following papers : 

Ophthalmic Lenses. 

Dioptric Formulse for Combined Cylindrical Lenses. 

The Prism=Dioptry. 

A Metric System of Numbering and Measuring Prisms. 

The Relation of the Prism-Dioptry to the Meter Angle. 

The Relation of the Prism-Dioptry to the Dens Dioptry. 

The Perfected Prismometer. 

The Prismometric Scale. 

On the Practical Execution of Ophthalmic Prescriptions involving Prisms. 
A Problem in Cemented Bifocal Lenses, Solved by the Prism=Dioptry. 

Why Strong Contra=Generic Lenses of Equal Power Fail to Neutralize 
Each Other. 

The Advantages of the Sphero=Toric Lens. 

The Iris, as Diaphragm and Photostat. 

The Typoscope. 

The Correction of Depleted Dynamic Refraction (Presbyopia). 


Press Notices on the Original Edition: 


OPHTHALMIC LENSES. 


"The work stands alone, in its present 
form, a compendium of the various laws of 
physics relative to this subject that are so 
difficult of access in scattered treatises.”— 
New England Medical Gazette. 

"It is the most complete and best illus¬ 
trated book on this special subject ever pub¬ 
lished.”— Horological Review, New York. 


" Of all the simple treatises on the prop¬ 
erties of lenses that we have seen, this is in¬ 
comparably the best. . . . The teacher of 
the average medical student will hail this 
little work as a great boon.”— Archives of 
Ophthalmology , edited byH. Knapp, M.D. 


DIOPTRIC FORMULA FOR COMBINED CYLINDRICAL LENSES. 


" This little brochure solves the problem 
of combined cylinders in all its aspects, and 
in a manner simple enough for the compre¬ 
hension of the average student of ophthal¬ 
mology. The author is to be congratulated 
upon the success that has crowned his labors, 
for nowhere is there to be found so simple 
and yet so complete an explanation as is con¬ 
tained in these pages.”— Archives of Oph¬ 
thalmology , edited by H. Knapp , M.D. 


"This exhaustive work of Mr. Prentice 
is a solution of one of the most difficult prob¬ 
lems in ophthalmological optics. Thanks 
are due to Mr. Prentice for the excellent 
manner in which he has elucidated a sub¬ 
ject which has not hitherto been satisfactor¬ 
ily explained.”— The Ophthalmic Review , 
London. 


The book contains 110 Original Diagrams. Bound in cloth. 


Price, $1.50 (6s. 3d.) 


Published by The Keystone, 

THE ORGAN OF THE JEWELRY AND OPTICAL TRADES, 
19TH & Brown Sts., Philadelphia, U. S. A. 






TESTS AND STUDIES OF THE 
OCULAR MUSCLES 

By Ernest E. Maddox, M. D., F. R. C. S., Ed 

Ophthalmic burgeon to the Royal Victoria Hospital, Bournemouth; formerly assistant 
ophthalmic surgeon to the Royal Infirmary, Edinburgh 


SECOND EDITION 

Revised and enlarged by the author 


The sub-division of refractive work that most troubles the 
optician is muscular anomalies. Even those who have mastered 
all the other intricacies of visual correction will often find their 
skill frustrated and their efforts nullified if they have not 
thoroughly mastered the ocular muscles—their functions and 
failings. The optician can thoroughly equip himself in this 
fundamental essential by studying the excellent treatise “The 
Ocular Muscles,” by Dr. Maddox, who is recognized in the 
medical world as one of the leading authorities on the subject 
of eye muscles. 

This work is the most complete and masterful ever com¬ 
piled on this important branch of ophthalmology, covering 
thoroughly the symptoms, tests and treatment of muscular 
anomalies. The accomplished author has devoted a lifetime of 
study and research to his subject, and the book throughout is 
marked by an explicitness and simplicity of language that make 
its study a pleasure to the eye specialist. 

Bound in Cloth. Over 100 Well Executed Illustrations 
Sent postpaid to any part of the world on receipt of SI.50 (6s. 3d.) 


Published by The Keystone 

THE ORGAN OF T 1 IE JEWELRY AND OPTICAL TRADES 
19TH & Brown Sts., Philadelphia, U. S. A. 





Optometric Record Boor 


A record book, wherein to record optometric examinations, 
is an indispensable adjunct of an optician’s outfit. 

The Keystone Optometric Record Book was specially pre¬ 
pared for this purpose. It excels all others in being not only a 
record book, but an invaluable guide in examination. 

The book contains two hundred record forms with printed 
headings, suggesting, in the proper order, the course of examina¬ 
tion that should be pursued to obtain most accurate results. 

Each book has an index, which enables the optician to refer 
instantly to the case of any particular patient. 

The Keystone Record Book diminishes the time and labor 
required for examinations, obviates possible oversights from 
carelessness and assures a systematic and thorough examination 
of the eye, as well as furnishes a permanent record of all exam¬ 
inations. 

Sent postpaid on receipt of 85 f .OO (4s. 2d.) 


Published by I HE K.EYSTONE, 

THE ORGAN OF THE JEWELRY AND OPTICAL TRADES, 
19TH & Brown Sts., Philadelphia, U.S.A. 









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